Class 10 Chemistry Chapter 5 Mole Concepts and Stoichimetry

Key Concept: STP Definition

c) $0^\circ \text{C}$ and $1 \, \text{atm}$
[Solution Description]
Standard Temperature Pressure (STP) is defined as $0^\circ \text{C}$ or $273 \, \text{K}$ for temperature and $1$ atmospheric pressure ($760 \, \text{mm}$ of $\text{Hg}$) for pressure.

Key Concept: Empirical Formula, Percentage Composition

a) CH$_2$Cl
[Solution Description]
To find the empirical formula:
1. Assume 100 g of the compound: Carbon = 24.27 g, Hydrogen = 4.07 g, Chlorine = 71.65 g.
2. Convert grams to moles:
Moles of Carbon = $\frac{24.27}{12} = 2.0225$ mol
Moles of Hydrogen = $\frac{4.07}{1} = 4.07$ mol
Moles of Chlorine = $\frac{71.65}{35.5} = 2.0183$ mol
3. Divide each mole value by the smallest number of moles (2.0183):
C = $\frac{2.0225}{2.0183} \approx 1$, H = $\frac{4.07}{2.0183} \approx 2$, Cl = $\frac{2.0183}{2.0183} = 1$
4. The simplest whole number ratio is C:H:Cl = 1:2:1.

Key Concept: Boyle's Law

c) Pressure halves
[Solution Description]
According to Boyle's Law, $P_1V_1 = P_2V_2$ at constant temperature. If the volume ($V$) is doubled ($V_2 = 2V_1$), then:
$P_1V_1 = P_2(2V_1)$
Solving for $P_2$:
$P_2 = \frac{P_1V_1}{2V_1} = \frac{P_1}{2}$
The pressure decreases by half.

Key Concept: Charles's Law, Volume Change

c) 91.4 mL
[Solution Description]
Using Charles’s Law, first convert temperatures to Kelvin: $T_1 = 0^\circ C + 273 = 273 K$ and $T_2 = 100^\circ C + 273 = 373 K$.
The initial volume $V_1 = 250 \, \text{mL}$. According to Charles’s Law, $\frac{V_1}{T_1} = \frac{V_2}{T_2}$, so $V_2 = V_1 \times \frac{T_2}{T_1} = 250 \times \frac{373}{273} \approx 341.39 \, \text{mL}$.
The change in volume is $V_2 - V_1 = 341.39 - 250 = 91.39 \, \text{mL}$.

Key Concept: Empirical Formula, Percentage Composition

b) 30.45\%
[Solution Description]
Step 1: Write the general decomposition reaction:
MCO$_3$ → MO + CO$_2$.
Step 2: Calculate moles of CO$_2$ produced = $\frac{1.25\, \text{g}}{44\, \text{g/mol}} = 0.0284$ mol.
Step 3: From stoichiometry, moles of MCO$_3$ = moles of CO$_2$ = 0.0284 mol.
Step 4: Calculate molar mass of MCO$_3$ = $\frac{2.45\, \text{g}}{0.0284\, \text{mol}} = 86.27$ g/mol.
Step 5: Subtract mass of CO$_3^{2-}$ group (60 g/mol) to find atomic mass of M = 86.27 - 60 = 26.27 g/mol (≈ Mg).
Step 6: Calculate \% metal = $\frac{24}{84} \times 100 = 28.57\%$ (using MgCO$_3$ as example).
Correction: Actual calculation should be $\frac{26.27}{86.27} \times 100 = 30.45\%$.

Key Concept: Vapour Density

c) 32 g/mol
[Solution Description]
Molecular mass is twice the vapour density: $2 \times 16 = 32 \text{ g/mol}$.

Key Concept: Mole concept, molar volume

d) 56 g and $12.044 \times 10^{23}$ molecules
[Solution Description]
Solution has two parts:
1) Mass calculation:
At STP, 1 mole of any gas occupies 22.4 L
Given volume is 44.8 L which corresponds to $\frac{44.8}{22.4} = 2$ moles
For $N_2$:
Molar mass = 28 g/mol
So, mass of 2 moles $N_2 = 2 \times 28 = 56$ g
2) Number of molecules:
2 moles of $N_2$ will contain $2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23}$ molecules
Combining both results gives the correct option.

Key Concept: Empirical and Molecular Formula Relationship

b) $C_3H_6O_3$
[Solution Description]
The empirical formula weight of $CH_2O$ is calculated as:
Carbon: $12$ g/mol, Hydrogen: $1 \times 2 = 2$ g/mol, Oxygen: $16$ g/mol.
Total empirical weight = $12 + 2 + 16 = 30$ g/mol.
Given $n = \frac{\text{Molecular Weight}}{\text{Empirical Weight}} = \frac{90}{30} = 3$.
Therefore, the molecular formula is $C_3H_6O_3$.

Key Concept: Percentage composition, Alloy analysis

b) 11.76\%
[Solution Description]
Consider 100 g of alloy:
Mass of Cu = 80 g → Moles of Cu = $80/63.5 = 1.26$ mol
Mass of Sn = 20 g → Moles of Sn = $20/118.7 = 0.168$ mol
Total moles = $1.26 + 0.168 = 1.428$ mol
\% of Sn atoms = $(0.168/1.428) \times 100 = 11.76\%$

Key Concept: Percentage Composition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify Assertion (A), we first calculate the molecular weight of $\text{H}_2\text{O}$:
$\text{Molecular weight of } \text{H}_2\text{O} = (2 \times 1) + 16 = 18$
The total weight of oxygen in one molecule is 16 g/mol.
Now, apply the percentage composition formula for oxygen:
$\text{Percentage of O} = \frac{16}{18} \times 100 = 88.89\%$
Thus, Assertion (A) is true. Reason (R) states the correct formula for calculating percentage composition. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Empirical and Molecular Formulas

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the molecular formula is always an integer multiple of the empirical formula, which is true because the molecular formula can be written as $\text{Empirical Formula} \times n$, where $n$ is obtained from the ratio of molecular weight to empirical formula weight. The reason correctly explains the assertion by stating that $n$ is calculated as $\frac{\text{Molecular Weight}}{\text{Empirical Formula Weight}}$. Since both are true and the reason correctly explains the assertion, the correct answer is option (a).

Key Concept: Statistical Normality, Central Limit Theorem

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion is false because while the sample mean converges to the population mean ($\mu$) as sample size increases, it does not "always" equal $\mu$ for finite samples. The Reason is true and independently explains the behavior of sample means under CLT, but it does not justify the Assertion.

Key Concept: Limiting Reagent

c) Both are limiting reagents
[Solution Description]
Balanced equation: $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$.
Moles of C$_2$H$_4$ = $\frac{28}{28} = 1$ mole.
Moles of O$_2$ = $\frac{96}{32} = 3$ moles.
According to the equation, 1 mole of C$_2$H$_4$ reacts with 3 moles of O$_2$.
Here, the reactants are present in the exact stoichiometric ratio (1:3).
Thus, no reactant is in excess; both are equally limiting in this case.

Key Concept: Balancing Reactions

c) 16 grams
[Solution Description]
The balanced chemical equation for combustion of methane is:
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
Applying the Law of Conservation of Mass, the mass of reactants (methane + oxygen) must equal the mass of products (carbon dioxide + water). Since 16 grams of $CH_4$ reacts, and assuming sufficient oxygen, the total mass of products will also be 16 grams (as mass cannot be created or destroyed).

Key Concept: Limiting Reagent, Excess Reagent

d) Assertion is false, but Reason is true.
[Solution Description]
We analyze the given reaction $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.
Step 1: Calculate the moles of methane and oxygen using ideal gas assumptions (since volumes are given at same conditions). 80 cm$^3$ of methane is 0.08 L, and 200 cm$^3$ of oxygen is 0.2 L. Under same conditions, volume ratios are equal to mole ratios, so we have 0.08 moles of methane and 0.2 moles of oxygen.
Step 2: Check stoichiometry. For every 1 mole of methane, 2 moles of oxygen are required. Thus, 0.08 moles of methane require 0.16 moles of oxygen. Since 0.2 moles of oxygen are available, oxygen is in excess, and methane is the limiting reagent.
Therefore, the assertion that oxygen is the limiting reagent is false. However, the reason correctly states the stoichiometric requirement (2 moles of oxygen per mole of methane), which is true. Hence, the correct option is d).

Key Concept: Molar mass, Mole concept

c) 2
[Solution Description]
The molar mass of $H_2SO_4$ is:
Hydrogen ($H$): $2 \times 1 = 2$ g/mol
Sulfur ($S$): $1 \times 32 = 32$ g/mol
Oxygen ($O$): $4 \times 16 = 64$ g/mol
Total molar mass: $2 + 32 + 64 = 98$ g/mol
49 grams of $H_2SO_4$ is half a mole because $49/98 = 0.5$ moles.
Each mole of $H_2SO_4$ contains 4 moles of oxygen atoms.
So, 0.5 moles of $H_2SO_4$ contain $0.5 \times 4 = 2$ moles of oxygen atoms.

Key Concept: Limiting Reactant, Excess Reactant

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, write the balanced chemical equation: $2H_2 + O_2 \rightarrow 2H_2O$.
Given moles of $H_2 = 4$ and $O_2 = 3$.
For $H_2$, the required moles of $O_2$ based on stoichiometry would be $4/2 = 2$ moles.
Since 3 moles of $O_2$ are available, which is more than the required 2 moles, $H_2$ is indeed the limiting reactant.
The reason states that the limiting reactant has a smaller mole ratio compared to its stoichiometric coefficient. Here, for $H_2$, the available mole ratio ($4/2 = 2$) is less than that of $O_2$ ($3/1 = 3$). Thus, the reason correctly explains why $H_2$ is the limiting reactant.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Mole Concept

b) 2 moles
[Solution Description]
First, calculate the molar mass of $CO_2$:
$\text{Molar Mass} = 12 + (2 \times 16) = 44 \text{ g/mol}$
Now, find the number of moles using:
$\text{Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{88}{44} = 2 \text{ moles}$

Key Concept: Avogadro’s Law

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that one mole of any gas at STP occupies 22.4 litres, which is correct according to the mole concept. The reason states Avogadro’s number as $6.022 \times 10^{23}$ particles per mole, which is also true. However, the reason does not explain why one mole of gas occupies 22.4 litres at STP; it merely states a different fact about Avogadro’s number. Thus, both are true, but the reason does not correctly explain the assertion.

Key Concept: Molar mass, Mole concept

b) $2.16792 \times 10^{25}$
[Solution Description]
First, calculate the molar mass of $Al_2(SO_4)_3 \cdot 18H_2O$:
Aluminum ($Al$): $2 \times 27 = 54$ g/mol
Sulfur ($S$): $3 \times 32 = 96$ g/mol
Oxygen ($O$) from sulfate: $12 \times 16 = 192$ g/mol
Water ($H_2O$): $18 \times (2 + 16) = 324$ g/mol
Total molar mass: $54 + 96 + 192 + 324 = 666$ g/mol
666 grams of the compound = 1 mole.
Number of water molecules in 1 mole: $18 \times 6.022 \times 10^{23}$.
Each $H_2O$ molecule contains 2 hydrogen atoms.
Total hydrogen atoms: $2 \times 18 \times 6.022 \times 10^{23} = 2.16792 \times 10^{25}$.

Key Concept: Identifying Balanced Equation

b) $P_4 + 5O_2 \rightarrow P_4O_{10}$
[Solution Description]
Verify each option:
1) $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ is balanced (C:1, H:4, O:4 on both sides).
2) $Fe + 2HCl \rightarrow FeCl_2 + H_2$ is balanced (Fe:1, H:2, Cl:2 on both sides).
3) $N_2 + 3H_2 \rightarrow 2NH_3$ is balanced (N:2, H:6 on both sides).
4) $2Na + Cl_2 \rightarrow 2NaCl$ is balanced (Na:2, Cl:2 on both sides).
All options are balanced except the distractors in the choices below.

Key Concept: Standard Temperature Pressure (STP

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that at STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 liters. This is a fundamental concept derived from Avogadro's law under standard conditions. The reason provided correctly defines the conditions of STP as 273 K (0°C) and 1 atmospheric pressure (760 mm Hg). Since the reason accurately explains why the assertion holds true (as 22.4 L/mol is defined under these exact conditions), both statements are correct, and the reason is indeed the correct explanation for the assertion.

Key Concept: Molar Volume Calculation

c) 4.48 L
[Solution Description]
First, find the molecular mass of $N_2$:
$2 \times 14 = 28$ g/mol.
28 g of $N_2$ occupies 22.4 L at STP.
For 5.6 g:
$\frac{22.4 \times 5.6}{28} = 4.48$ L.

Key Concept: Avogadro's Number, Molecular Mass, Moles

c) 44 g
[Solution Description]
The molecular mass of $CO_2$ is calculated as follows:
Carbon ($C$) has a molar mass of 12 g/mol, and Oxygen ($O$) has a molar mass of 16 g/mol. Since $CO_2$ has one carbon atom and two oxygen atoms, its molecular mass is:
$12 + (2 \times 16) = 44 \text{ g/mol}$
Avogadro's number ($6.022 \times 10^{23}$) represents one mole of any substance. Therefore, the mass of $6.022 \times 10^{23}$ molecules of $CO_2$ is equal to its molar mass, which is 44 g.

Key Concept: Mole Concept

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the molar volume of any gas at S.T.P. is 22.4 litres, which is correct as per Avogadro's Law. The reason explains that one mole of any gas at S.T.P. contains Avogadro’s number of molecules ($6.022 \times 10^{23}$) and occupies 22.4 litres, which is also correct. Moreover, the reason correctly explains the assertion by linking the concept of molar volume to Avogadro’s number. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Molecular Formula Calculation

d) $\text{C}_6\text{H}_{12}\text{O}_6$
[Solution Description]
First, calculate the empirical formula weight:
Empirical formula weight = $12 \, (\text{C}) + 2 \times 1 \, (\text{H}) + 16 \, (\text{O}) = 30$.
Now, find $n$ using the formula $n = \frac{\text{Molecular Weight}}{\text{Empirical Formula Weight}}$:
$n = \frac{180}{30} = 6$.
Multiply the empirical formula by $n$:
Molecular formula = $\text{C}_6\text{H}_{12}\text{O}_6$.

Key Concept: Molar Volume and STP

b) 5.6 litres
[Solution Description]
The molar mass of $N_2$ is $2 \times 14 = 28$ g/mol. The number of moles in 7 g of $N_2$ is:
$\text{Moles} = \frac{7}{28} = 0.25 \text{ moles}.$
At STP, 1 mole of any gas occupies 22.4 litres. Therefore, 0.25 moles will occupy:
$0.25 \times 22.4 = 5.6 \text{ litres}.$

Key Concept: Gram Molecular Mass

c) 17 g/mol
[Solution Description]
The molecular formula of ammonia is $NH_3$. To find its molar mass, we add the atomic masses of nitrogen and hydrogen. Given: $N = 14$, $H = 1$.
Molar mass of $NH_3 = 14 + (3 \times 1) = 17$ g/mol.

Key Concept: Gram Molecular Mass

d) 32 g
[Solution Description]
The molecular mass of oxygen gas ($O_2$) is 32 u (16 u × 2). Therefore, one mole of $O_2$ weighs equal to its gram molecular mass, which is 32 g.

Key Concept: Moles and Particles

b) $3.011 \times 10^{23}$
[Solution Description]
Given moles of $NH_3 = 0.5$. Using the relationship between moles and particles:
$\text{Number of particles} = \text{Moles} \times 6.022 \times 10^{23} = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \, \text{molecules}$
Thus, 0.5 moles of $NH_3$ contain $3.011 \times 10^{23}$ molecules.

Key Concept: Boyle’s Law, combined concepts with unit conversion

a) 2.5 L
[Solution Description]
Using Boyle’s Law: $P_1V_1 = P_2V_2$, where:
- $P_1 = 760$ mmHg, $V_1 = 5$ L,
- $P_2 = 1520$ mmHg, $V_2 = ?$.
Substituting:
$760 \times 5 = 1520 \times V_2$
Divide both sides by 1520:
$V_2 = \frac{3800}{1520} = 2.5 \, \text{L}$
The new volume is 2.5 L.

Key Concept: Gas Laws and Stoichiometry

c) 2.8 L
[Solution Description]
Molar mass of $\text{SO}_2$ = $32 + 2(16) = 64$ g/mol.
Moles of $\text{SO}_2$ = $\frac{8}{64} = 0.125$ mol.
At STP, 1 mole of any gas occupies 22.4 L.
Volume of $\text{SO}_2$ = $0.125 \times 22.4 = 2.8$ L.

Key Concept: Atomicity Calculation

b) 2
[Solution Description]
First, convert the given volume to litres:
$11,200 \text{ cm}^3 = 11,200 \times 10^{-3} \text{ L} = 11.2 \text{ L}$
Calculate the number of moles of chlorine gas:
$\text{Number of moles} = \frac{\text{Volume}}{\text{Molar volume}} = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ mol}$
Now, find the molar mass of chlorine:
$\text{Molar mass} = \frac{\text{Mass}}{\text{Number of moles}} = \frac{35.5 \text{ g}}{0.5 \text{ mol}} = 71 \text{ g/mol}$
Since the atomic mass of chlorine is 35.5, the atomicity ($n$) is:
$n = \frac{\text{Molecular mass}}{\text{Atomic mass}} = \frac{71}{35.5} = 2$

Key Concept: Determining Empirical Formula from Percentage Composition

a) $CH_2O$
[Solution Description]
Assume 100 g of the compound.
Moles of carbon = $\frac{40}{12} \approx 3.33$.
Moles of hydrogen = $\frac{6.7}{1} = 6.7$.
Moles of oxygen = $\frac{53.3}{16} \approx 3.33$.
Simplifying (divide by smallest value, 3.33):
C: $\frac{3.33}{3.33} = 1$, H: $\frac{6.7}{3.33} \approx 2$, O: $\frac{3.33}{3.33} = 1$.
Thus, the empirical formula is $CH_2O$.

Key Concept: STP Conditions

b) $0^\circ C$ and 1 atm
[Solution Description]
The standard values chosen are:
- Temperature: $0^\circ C$ or $273 K$
- Pressure: $1$ atmospheric pressure or $760$ mm or $76$ cm of Hg

Key Concept: Molecular Formula

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The molecular formula of water is indeed $\text{H}_2\text{O}$, which matches its empirical formula because the simplest ratio of hydrogen to oxygen here is already 2:1. Hence, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the relative molecular mass of a gas is twice its vapour density, which is correct as per the formula $2 \times \text{Rel. V.D.} = \text{Rel. molecular mass of a gas or vapour}$. The reason correctly explains the assertion by referring to Avogadro's Law, which allows substituting volumes with molecules and deriving the relationship between relative vapour density and molecular mass. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Avogadro's Law, Molecular Formula

a) He
[Solution Description]
Vapour density (VD) is half of the molecular mass (M). So, M = 2 × VD = 2 × 16 = 32.
Given that the atomic mass of the element is 32, the molecular formula must be monatomic. Hence, the molecular formula is same as the atomic symbol.

Key Concept: Mass Calculation of Excess Reactant

d) 0 g
[Solution Description]
1. Moles of $H_2$ = $\frac{10 \text{ g}}{2 \text{ g/mol}} = 5$ moles.
2. Moles of $O_2$ = $\frac{64 \text{ g}}{32 \text{ g/mol}} = 2$ moles.
3. From the equation, 2 moles of $H_2$ react with 1 mole of $O_2$.
4. So, 5 moles of $H_2$ need $\frac{5}{2} = 2.5$ moles of $O_2$.
5. Since only 2 moles of $O_2$ are available, $O_2$ is the limiting reactant.
6. No $O_2$ remains unreacted as all 2 moles are consumed.

Key Concept: Number of Molecules, Molar Volume

a) $9.033 \times 10^{21}$
[Solution Description]
Convert volume to litres: $336 \text{ cm}^3 = 0.336 \text{ dm}^3$. Moles of $O_2$ = $\frac{0.336}{22.4} = 0.015 \text{ moles}$. Number of molecules = moles × Avogadro's number = $0.015 \times 6.022 \times 10^{23} = 9.033 \times 10^{21}$.

Key Concept: Atomicity, Avogadro's Law

b) $3.011 \times 10^{23}$
[Solution Description]
22.4 L of any gas at STP contains $6.022 \times 10^{23}$ molecules (Avogadro's number). Therefore, 11.2 L will contain half of that, i.e., $\frac{6.022 \times 10^{23}}{2} = 3.011 \times 10^{23}$ molecules.

Key Concept: Boyle's Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that doubling the pressure halves the volume for a fixed mass of gas at constant temperature, which is true as per Boyle's Law ($P_1V_1 = P_2V_2$). The reason correctly cites Boyle's Law, and it explains why the assertion is true. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Gas Equation

c) 44.8 L
[Solution Description]
STP conditions are $T_1 = 273\,K$, $P_1 = 1\,atm$, $V_1 = 22.4\,L$. Using the combined gas law $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$, and since pressure is constant ($P_1 = P_2$):
$\frac{V_1}{T_1} = \frac{V_2}{T_2}$
Given $T_2 = 546\,K$:
$\frac{22.4}{273} = \frac{V_2}{546}$
Solving for $V_2$:
$V_2 = \frac{22.4 \times 546}{273} = 44.8\,L$

Key Concept: Drug Formulation

c) To ensure proper drug delivery and effectiveness
[Solution Description]
Excipients are inactive substances used as carriers for the active ingredients in medications. They serve multiple purposes including enhancing stability, improving bioavailability, and facilitating drug delivery. The primary purpose is to ensure proper drug delivery and effectiveness.

Key Concept: Gay-Lussac’s Law - Reaction of Hydrogen and Chlorine

c) 40 cm$^3$
[Solution Description]
The balanced equation is $H_2 + Cl_2 \rightarrow 2HCl$, indicating a volume ratio of 1:1:2 for $H_2$, $Cl_2$, and $HCl$ respectively.
Given:
- Volume of $H_2$ = 20 cm$^3$
- Volume of $Cl_2$ = 20 cm$^3$
Since the ratio is 1:1:2, the volume of $HCl$ produced will be twice the volume of either $H_2$ or $Cl_2$:
$2 \times 20 = 40 \text{ cm}^3.$

Key Concept:

c) 7
[Solution Description]
The balanced chemical equation for the combustion of ethane is:
$2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$
To balance carbon atoms, we require 2 molecules of $C_2H_6$ producing 4 molecules of $CO_2$. For hydrogen, 12 hydrogen atoms (from 2 molecules of $C_2H_6$) yield 6 molecules of $H_2O$. Lastly, to balance oxygen atoms, 14 oxygen atoms are required on the reactant side ($7O_2$).

Key Concept: Law of Conservation of Mass, Chemical Reactions

c) 20 grams
[Solution Description]
The reaction is:
$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$
Initial mass = $50\,\text{g}$. Final mass = $30\,\text{g}$ (excluding escaped $CO_2$).
By the Law of Conservation of Mass, the mass lost must be the mass of $CO_2$ that escaped.
So, mass of $CO_2$ = $50\,\text{g} - 30\,\text{g} = 20\,\text{g}$.

Key Concept: Avogadro’s Law and Mole Concept

c) $6.022 \times 10^{23}$ molecules
[Solution Description]
Avogadro's Law states that one mole of any substance contains $6.022 \times 10^{23}$ molecules (Avogadro's number). This applies to all substances, whether they are elements or compounds.

Key Concept: Mass-mass relationship, stoichiometry, balancing equations

b) 11.2 g
[Solution Description]
Molar mass of $CaCO_3 = 40 + 12 + 3 \times 16 = 100\,g/mol$.
Moles of $CaCO_3 = \frac{20\,g}{100\,g/mol} = 0.2\,mol$.
From the balanced equation, 1 mole of $CaCO_3$ produces 1 mole of $CaO$.
Thus, 0.2 moles of $CaCO_3$ produce 0.2 moles of $CaO$.
Molar mass of $CaO = 40 + 16 = 56\,g/mol$.
Mass of $CaO$ produced = $0.2\,mol \times 56\,g/mol = 11.2\,g$.

Key Concept: Percentage composition

b) 16.0\%
[Solution Description]
First, calculate the molecular mass of calcium carbonate:
$\text{Molecular mass of CaCO}_3 = 40 + 12 + (16 \times 3) = 100$
Mass of calcium in $\text{CaCO}_3$ per 100 g of sample:
$\text{Ca in CaCO}_3 = \frac{40}{100} \times 40 = 16$
Silica ($\text{SiO}_2$) does not contain calcium, so its contribution is 0.
Total calcium percentage in the sample:
$\text{Ca percentage} = 16\%$

Key Concept: Atomicity of Gases

b) 2
[Solution Description]
Hydrogen gas ($H_2$) is diatomic, meaning each molecule consists of 2 hydrogen atoms bonded together.

Key Concept: Calculating Percentage Composition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the percentage composition of carbon in glucose ($C_6H_{12}O_6$), we first determine the molecular mass:
Molecular mass of $C_6H_{12}O_6 = (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180$ g/mol.
The total mass of carbon in one molecule is $6 \times 12 = 72$ g.
Using the percentage composition formula:
$\text{Percentage of C} = \frac{72}{180} \times 100 = 40\%$.
Thus, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Avogadro's Number

b) $6.022 \times 10^{23}$
[Solution Description]
Avogadro's number ($N_A$) is defined as the number of elementary particles (atoms, molecules, ions) present in one mole of a substance. Its value is $6.022 \times 10^{23}$.

Key Concept: Counting Atoms

c) 2 on both sides
[Solution Description]
Count the oxygen atoms:
- Left Side (Reactants): $O_2$ has 2 oxygen atoms.
- Right Side (Products): $2H_2O$ has 2 molecules, each with 1 oxygen atom, totaling 2 oxygen atoms.
Both sides have 2 oxygen atoms.

Key Concept: Mole Concept, Avogadro’s Number, Mole of Molecules

b) $4.52 \times 10^{24}$
[Solution Description]
First, find the moles of water:
Molar mass of $H_2O = 2 \times 1 + 16 = 18$ g/mol.
Moles of water = $\frac{45 \text{ g}}{18 \text{ g/mol}} = 2.5$ moles.
Number of molecules in 2.5 moles = $2.5 \times 6.022 \times 10^{23} = 1.5055 \times 10^{24}$ molecules.
Each molecule of $H_2O$ contains 3 atoms (2 H + 1 O), so total atoms = $3 \times 1.5055 \times 10^{24} = 4.5165 \times 10^{24}$ atoms.

Key Concept: Combustion Reactions, Volume Relationships in Reactions

d) 150 cm$^3$
[Solution Description]
The balanced equation is:
$C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$
1 volume of $C_2H_4$ requires 3 volumes of $O_2$.
For 50 cm$^3$ of $C_2H_4$, the required $O_2$ volume is:
$50 \times 3 = 150$ cm$^3$.

Key Concept: Formula Application

d) $N = \frac{\text{equivalents of solute}}{\text{liters of solution}}$
[Solution Description]
Normality ($N$) is calculated using the following formula:
$N = \frac{\text{equivalents of solute}}{\text{liters of solution}}$
It represents the number of equivalents per liter of solution.

Key Concept: Gram molecular mass, Mole concept

a) 873.25 g
[Solution Description]
First, find the molar mass of $CuSO_4 \cdot 5H_2O$.
Molar mass = $63.5 + 32 + (16 \times 4) + 5(1 \times 2 + 16) = 63.5 + 32 + 64 + 90 = 249.5$ g/mol.
Now, multiply the number of moles by the molar mass: $3.5 \times 249.5 = 873.25$ g.

Key Concept: Avogadro's Number

c) $1.2044 \times 10^{24}$
[Solution Description]
First, find the molecular mass of $H_2O$:
Atomic mass of hydrogen (H) = 1 g/mol
Atomic mass of oxygen (O) = 16 g/mol
Molecular mass of $H_2O$ = $(2 \times 1) + 16 = 18$ g/mol
Now, calculate the number of moles of $H_2O$ in 36 g:
Moles = $\frac{\text{Mass}}{\text{Molecular Mass}} = \frac{36}{18} = 2$ moles
Since 1 mole contains $6.022 \times 10^{23}$ molecules, 2 moles contain:
$2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}$ molecules.

Key Concept: Avogadro's Number, Molecular Calculation

b) $6.022 \times 10^{22}$
[Solution Description]
Step 1: Calculate molar mass of H$_2$O = 2(1) + 16 = 18 g/mol
Step 2: Calculate moles of water:
$\text{moles} = \frac{1.8}{18} = 0.1 \text{ moles}$
Step 3: Calculate molecules of water:
$\text{molecules} = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$
Step 4: Each molecule has 1 oxygen atom, so total oxygen atoms = $6.022 \times 10^{22}$

Key Concept: Molar Volume

b) 22.4 L
[Solution Description]
According to Avogadro's Law, one mole of any gas occupies a volume of 22.4 litres at Standard Temperature and Pressure (STP). This is known as the molar volume.
Molar volume $= 22.4 \text{ dm}^3$ or $22.4 \text{ L}$

Key Concept: Mole Concept, Avogadro's Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
- **Step 1:** Calculate the number of moles in 18 g of $H_2O$.
Molar mass of $H_2O = 2(1) + 16 = 18$ g/mol.
Moles of $H_2O = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{18\,\text{g}}{18\,\text{g/mol}} = 1$ mole.
Number of molecules in 1 mole of $H_2O = 6.022 \times 10^{23}$.
- **Step 2:** Calculate the number of moles in 16 g of $O_2$.
Molar mass of $O_2 = 2(16) = 32$ g/mol.
Moles of $O_2 = \frac{16\,\text{g}}{32\,\text{g/mol}} = 0.5$ mole.
Number of molecules in 0.5 mole of $O_2 = 3.011 \times 10^{23}$.
- **Step 3:** Each $O_2$ molecule contains 2 atoms.
Total number of atoms in 16 g of $O_2 = 2 \times 3.011 \times 10^{23} = 6.022 \times 10^{23}$.
- **Conclusion:** Assertion is true as both quantities contain $6.022 \times 10^{23}$ entities. Reason is also true because it correctly explains why the assertion holds—both amounts correspond to one mole when considering molecular vs. atomic count.

Key Concept: Gay-Lussac’s Law - Unreacted Gas Volume

b) 25 cm\textsuperscript{3}
[Solution Description]
From the equation, $2H_2 + O_2 \rightarrow 2H_2O$, 2 volumes of $H_2$ react with 1 volume of $O_2$. Here, 150 cm\textsuperscript{3} of $H_2$ requires $\frac{150}{2} = 75$ cm\textsuperscript{3} of $O_2$. Since initially there was 100 cm\textsuperscript{3} of $O_2$, the unreacted oxygen is $100 - 75 = 25$ cm\textsuperscript{3}.

Key Concept: Pharmacokinetics

b) The fraction of drug that reaches systemic circulation unchanged
[Solution Description]
Bioavailability refers to the fraction of an administered dose of a drug that reaches systemic circulation unchanged. It's a key parameter in determining dosage regimens. For intravenous administration, bioavailability is 100\% since the drug is directly introduced into the bloodstream.

Key Concept: Gay-Lussac's Law, Volume Ratio Calculations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The balanced chemical equation for the reaction is:
$N_2 + 3H_2 \rightarrow 2NH_3$.
According to Gay-Lussac’s Law, the volume ratio of $N_2$ to $NH_3$ is 1:2.
Given 120 cm$^3$ of $N_2$, the volume of $NH_3$ produced can be calculated as follows:
Since 1 volume of $N_2$ produces 2 volumes of $NH_3$,
120 cm$^3$ of $N_2$ will produce $120 \times 2 = 240$ cm$^3$ of $NH_3$.
Therefore, the Assertion (A) is true.
The Reason (R) states the correct volume ratio from Gay-Lussac’s Law, which directly explains why the volume of $NH_3$ is 240 cm$^3$.
Thus, both (A) and (R) are true, and (R) is the correct explanation of (A).

Key Concept: Empirical Formula, Water of Crystallisation

c) $MgSO_4 \cdot 7H_2O$
[Solution Description]
Convert percentages to moles:
$\text{Magnesium} = \frac{9.76}{24} \approx 0.407$
$\text{Sulfur} = \frac{13.01}{32} \approx 0.407$
$\text{Oxygen} = \frac{26.01}{16} \approx 1.626$
$\text{Water} = \frac{51.22}{18} \approx 2.846$
Divide each by the smallest value (0.407):
$\text{Magnesium} = 1$
$\text{Sulfur} = 1$
$\text{Oxygen} = 4$
$\text{Water} = 7$
Thus, the empirical formula is:
$MgSO_4 \cdot 7H_2O$

Key Concept: Temperature-Volume Relationship, Absolute Zero, STP

b) $\frac{1}{2}$ of the original volume
[Solution Description]
Convert the final temperature to Kelvin: $T_2 = -136.5 + 273 = 136.5 K$.
The initial temperature at STP is $T_1 = 273 K$.
Using Charles's Law:
$\frac{V_1}{T_1} = \frac{V_2}{T_2}$
Substitute the values:
$\frac{22.4}{273} = \frac{V_2}{136.5}$
Solve for $V_2$:
$V_2 = \frac{22.4 \times 136.5}{273} = 11.2 \, \text{L}$
Fraction of original volume:
$\frac{V_2}{V_1} = \frac{11.2}{22.4} = 0.5$

Key Concept: Excess and Limiting Reactants

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine the limiting reactant, compare the mole ratio of reactants with the stoichiometric ratio. For the given reaction, the stoichiometric ratio is $2:1$ ($H_2:O_2$). Here, the available mole ratio is $4:3$. Dividing the moles of each reactant by their stoichiometric coefficients gives $4/2 = 2$ for $H_2$ and $3/1 = 3$ for $O_2$. Since $2 < 3$, $H_2$ is the limiting reactant as it will be consumed first. Thus, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Identifying Unbalanced Elements

c) Hydrogen and oxygen
[Solution Description]
Before balancing, count the atoms on both sides for each element:
1. Phosphorus (P): 1 on left, 1 on right (balanced).
2. Hydrogen (H): 1 on left ($HNO_3$), 5 on right (3 in $H_3PO_4$ + 2 in $H_2O$) → unbalanced.
3. Nitrogen (N): 1 on left, 1 on right (balanced).
4. Oxygen (O): 3 on left, 7 on right (4 in $H_3PO_4$ + 2 in $NO_2$ + 1 in $H_2O$) → unbalanced.
Thus, hydrogen and oxygen are unbalanced.

Key Concept: Charles's Law

b) 600 mL
[Solution Description]
Using Charles's Law, $\frac{V_1}{T_1} = \frac{V_2}{T_2}$, where $V_1 = 300 \, \text{mL}$, $T_1 = 27^\circ \text{C} = 300 \, \text{K}$, and $T_2 = 327^\circ \text{C} = 600 \, \text{K}$. Solving for $V_2$:
$V_2 = \frac{V_1 \times T_2}{T_1} = \frac{300 \times 600}{300} = 600 \, \text{mL}.$
The volume doubles when absolute temperature doubles, resulting in 600 mL.

Key Concept: Molecular Formula from Empirical Formula

c) $C_6H_6$
[Solution Description]
1. Calculate the empirical formula weight of $CH$ = $12 + 1 = 13$ g/mol.
2. Compute $n$ = $\frac{\text{Molecular Weight}}{\text{Empirical Formula Weight}} = \frac{78}{13} = 6$.
3. Multiply the empirical formula by $n$: $CH \times 6 = C_6H_6$.
The molecular formula is $C_6H_6$.

Key Concept: Relative V.D. and Molar Volume

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Relative vapour density (Rel. V.D.) is related to molecular mass by the formula:
$2 \times \text{Rel. V.D.} = \text{Molecular mass}$
Given Rel. V.D. = 22,
Molecular mass = $2 \times 22 = 44$ g/mol.
Thus, both Assertion (A) and Reason (R) are true, and Reason correctly explains the Assertion.

Key Concept: Volume-based Limiting Reagent Problem

b) 100 cm³
[Solution Description]
The stoichiometric ratio is 1:3 for $N_2$:$H_2$. Here, both reactants are in the exact ratio (50:150 = 1:3), so neither is in excess. Volume of $NH_3$ formed = $2 \times 50 = 100$ cm³ (from the equation, 1 volume of $N_2$ gives 2 volumes of $NH_3$).

Key Concept: Relative Vapour Density, Molecular Mass

c) 46
[Solution Description]
Given the relationship between Relative Vapour Density (RVD) and Molecular Mass (M) is:
$M = 2 \times \text{RVD}$
Substituting RVD = 23,
$M = 2 \times 23 = 46$
Therefore, the molecular mass of the gas is 46.

Key Concept: Boyle's Law

b) 2.0 atm
[Solution Description]
Using Boyle's Law, $P_1V_1 = P_2V_2$, where $P_1 = 1.2 \, \text{atm}$, $V_1 = 2.5 \, \text{L}$, and $V_2 = 1.5 \, \text{L}$. Solving for $P_2$:
$P_2 = \frac{P_1V_1}{V_2} = \frac{(1.2)(2.5)}{1.5} = 2.0 \, \text{atm}.$
The new pressure is 2.0 atm.

Key Concept: Avogadro’s Number, Mole Concept

d) Assertion is false, but Reason is true.
[Solution Description]
To analyze the assertion and reason:
1) The molar mass of $O_2$ is 32 g/mol (16 g/mol × 2), while the molar mass of $O_3$ is 48 g/mol (16 g/mol × 3). Thus, the assertion is false since their masses differ.
2) The reason states that both gases contain the same number of molecules, which is true because 1 mole of any substance contains Avogadro's number ($6.022 \times 10^{23}$) of entities, regardless of the type of gas or its molecular weight. However, this does not explain the assertion, as the assertion itself is false.
Therefore, Assertion is false, but Reason is true.

Key Concept: Mole Concept, Molar Mass Calculations, Stoichiometric Problems

b) 44 g/mol, 0.5 moles
[Solution Description]
At STP, 1 mole of any gas occupies 22.4 L. Given volume = 11.2 L, so number of moles = $\frac{11.2}{22.4} = 0.5$ moles. Molar mass = $\frac{\text{Mass}}{\text{Moles}} = \frac{22}{0.5} = 44$ g/mol.

Key Concept: Gay-Lussac’s Law, Multiple Reactions

c) 100 mL
[Solution Description]
According to the reaction: 1 volume of $H_2S$ reacts with 1 volume of $Cl_2$ to produce 2 volumes of $HCl$.
Given: 50 mL $H_2S$ and 70 mL $Cl_2$. Here, $H_2S$ is the limiting reactant.
Volume of $HCl$ produced = $2 \times 50 = 100$ mL (since 1 mole of $H_2S$ gives 2 moles of $HCl$).

Key Concept: Gay-Lussac's Law of Combining Volumes and Avogadro's Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Gay-Lussac's law states that gases react in simple whole-number volume ratios. For the reaction $2H_2 + O_2 \rightarrow 2H_2O$, the volume ratio is indeed 2:1 for hydrogen to oxygen, making the Assertion true. Avogadro's law explains this by stating that equal volumes of gases contain the same number of molecules under identical conditions, which supports the observed ratio. Thus, both statements are true, and the Reason correctly explains the Assertion.

Key Concept: Excess Reactant Analysis

b) $O_2$ with 2 moles remaining
[Solution Description]
According to the balanced equation, 1 mole of $C_3H_8$ reacts with 5 moles of $O_2$. For 2 moles of $C_3H_8$, the required $O_2$ is $2 \times 5 = 10$ moles. Since 12 moles of $O_2$ are provided, $O_2$ is in excess by $12 - 10 = 2$ moles. Thus, $C_3H_8$ is completely consumed, and $O_2$ remains in excess by 2 moles.

Key Concept: Percentage Composition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Calculate the molecular weight of water.
Molecular weight of $\text{H}_2\text{O} = (2 \times 1 \text{g/mol}) + (1 \times 16 \text{g/mol}) = 18 \text{g/mol}$.
Step 2: Calculate the mass contribution of hydrogen and oxygen in water.
Mass of hydrogen = $2 \times 1 \text{g/mol} = 2 \text{g/mol}$.
Mass of oxygen = $1 \times 16 \text{g/mol} = 16 \text{g/mol}$.
Step 3: Compute the percentage composition.
Percentage of hydrogen = $\frac{2}{18} \times 100 \approx 11.11\%$.
Percentage of oxygen = $\frac{16}{18} \times 100 \approx 88.89\%$.
Step 4: Verify the mass ratio.
The ratio of hydrogen to oxygen mass is $2:16 = 1:8$.
Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Law of Conservation of Mass, Combustion

c) 40 grams
[Solution Description]
The balanced equation is:
$2H_2 + O_2 \rightarrow 2H_2O$
Molar mass of $H_2$ = $2\,\text{g/mol}$, $O_2$ = $32\,\text{g/mol}$.
For every 4 grams of $H_2$, 32 grams of $O_2$ is needed (ratio 1:8 by mass).
Here, $5\,\text{g}$ of $H_2$ requires $(5\,\text{g}) \times 8 = 40\,\text{g}$ of $O_2$.

Key Concept: Identifying the Limiting Reagent

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Write the balanced equation: $2SO_2 + O_2 \rightarrow 2SO_3$.
Step 2: The stoichiometric ratio is 2:1 ($SO_2:O_2$). For 5 moles of $SO_2$, the required moles of $O_2$ are $\frac{5}{2} = 2.5$ moles.
Step 3: Only 2 moles of $O_2$ are available, which is less than the required 2.5 moles. Thus, $O_2$ is indeed the limiting reagent.
Step 4: The Reason correctly explains why $O_2$ is the limiting reagent because it follows the stoichiometric constraint.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Atomicity determination

c) 4 litres
[Solution Description]
From the balanced equation, 2 volumes of $H_2$ produce 2 volumes of $H_2O$. Thus, the ratio is 1:1. So, 4 litres of $H_2$ will produce 4 litres of $H_2O$ vapor under same conditions.

Key Concept: Vapour Density and Molecular Formula

b) C$_4$H$_{10}$
[Solution Description]
Molecular Mass = 2 × Vapour Density = 2 × 29 = 58 g/mol.
Empirical formula mass of C$_2$H$_5$ = 2(12) + 5(1) = 29 g/mol.
$n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{58}{29} = 2$.
Molecular Formula = (C$_2$H$_5$)$_2$ = C$_4$H$_{10}$.

Key Concept: Mole Concept and Molar Mass Calculations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is true because the molar mass of a gas can indeed be calculated using its mass and volume at STP through the formula $\text{Molar mass} = \frac{\text{Mass}}{\text{Number of moles}}$, where the number of moles is found using $n = \frac{V}{22.4}$ at STP. The reason is also true as per Avogadro's law, which states that one mole of any gas occupies 22.4 L at STP. Moreover, the reason correctly explains why the assertion holds true, as it provides the fundamental principle behind the calculation method.

Key Concept: Gram Molecular Mass

b) 18 g
[Solution Description]
The gram molecular mass of a compound is equal to its molecular mass expressed in grams. Since the molecular mass of water is $18 \, \text{a.m.u.}$, its gram molecular mass is:
$18 \, \text{g}$

Key Concept: Dilution and Molarity

b) 0.4 M
[Solution Description]
Use the dilution formula: $M_1V_1 = M_2V_2$.
Here, $M_1 = 2 \text{M}$, $V_1 = 100 \text{mL}$, and $V_2 = 500 \text{mL}$.
Rearrange to find $M_2 = (M_1V_1)/V_2 = (2 \times 100)/500 = 0.4 \text{M}$.

Key Concept: Molarity Calculation

b) 1 M
[Solution Description]
To find the molarity, first calculate the number of moles of NaCl:
Moles of NaCl = Mass / Molar mass = $29.25 \text{g} / 58.5 \text{g/mol} = 0.5 \text{mol}$.
Next, convert the volume from mL to L: $500 \text{mL} = 0.5 \text{L}$.
Now, calculate molarity using the formula: Molarity = Moles of solute / Volume of solution in liters = $0.5 \text{mol} / 0.5 \text{L} = 1 \text{M}$.

Key Concept: Molar Mass of Water

c) 18 g
[Solution Description]
The gram molecular mass (molar mass) of water ($H_2O$) is given as 18 g/mol in the syllabus. Each molecule of water has 2 hydrogen atoms and 1 oxygen atom, which adds up to a molar mass of 18 g.

Key Concept: Percentage Composition

b) 27.3\%
[Solution Description]
The molecular mass of \$CO_2\$ is:
$12 + 2 \times 16 = 44$
Carbon contributes \$12\$ to the total molecular mass. The percentage of carbon is:
$\frac{12}{44} \times 100 = 27.3\%$
Thus, the percentage of carbon in \$CO_2\$ is approximately 27.3\%.

Key Concept: Percentage composition, Hydrated salts

c) 10
[Solution Description]
First calculate molecular weight of anhydrous $Na_2CO_3$:
$23 \times 2 + 12 + 16 \times 3 = 106$ g/mol.
Let x be the number of water molecules. Total mass of hydrate is:
$106 + x(18)$.
Given \% water = 62.9\%, so:
$\frac{18x}{106 + 18x} \times 100 = 62.9$.
Solving: $1800x = 62.9(106) + 1132.2x$
$667.8x = 6667.4$
$x = 10$

Key Concept: Molar Volume

b) 22.4 litres
[Solution Description]
According to the molar volume concept, 1 mole of any gas occupies 22.4 litres at STP. This is a standard value.

Key Concept: Normality

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Normality ($N$) is calculated as the number of gram equivalents of solute divided by the volume of solution in liters. The assertion correctly states that it varies with reaction conditions (e.g., acid-base reactions may change equivalents), and the reason defines normality accurately. Hence, the reason explains the assertion.

Key Concept: Atomicity of Nitrogen and Oxygen

b) 1
[Solution Description]
Given the reaction:
$N_2 + O_2 \rightarrow 2NO$
Half a molecule of $N_2$ is equivalent to 1 nitrogen atom, and half a molecule of $O_2$ is equivalent to 1 oxygen atom. They react to form 1 molecule of $NO$, as per the balanced equation.

Key Concept: Gay-Lussac’s Law, Volume Ratios

b) 200 mL
[Solution Description]
According to Gay-Lussac’s Law, gases react in simple whole number ratios by volume.
1 volume of $N_2$ reacts with 3 volumes of $H_2$ to produce 2 volumes of $NH_3$.
Given: 100 mL of $N_2$ reacts completely.
Volume of $H_2$ required = $3 \times 100 = 300$ mL.
Volume of $NH_3$ produced = $2 \times 100 = 200$ mL.
Total volume of gaseous mixture after reaction = Volume of $NH_3$ formed (since all reactants are consumed).
Thus, total volume = 200 mL.

Key Concept: Mass Calculations

c) 58.5 g
[Solution Description]
1 mole of Na (23 g) produces 1 mole of NaCl (23 + 35.5 = 58.5 g). Since the equation shows that 2 moles of Na produce 2 moles of NaCl, the ratio is 1:1. Therefore, 23 g of Na produces 58.5 g of NaCl.

Key Concept: Balancing Chemical Equations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Verify if the given chemical equation $2H_2 + O_2 \rightarrow 2H_2O$ is balanced.
- Left side: 4 H atoms and 2 O atoms.
- Right side: 4 H atoms and 2 O atoms.
Since the number of atoms of each element is equal, the equation is balanced.
Step 2: Check the Reason statement.
The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction, which requires balancing the equation to ensure equal numbers of atoms on both sides.
Conclusion: Both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Limiting Reagent

b) Hydrogen (H$_2$)
[Solution Description]
Balanced equation: $N_2 + 3H_2 \rightarrow 2NH_3$.
Moles of N$_2$ = $\frac{56}{28} = 2$ moles.
Moles of H$_2$ = $\frac{10}{2} = 5$ moles.
According to the equation, 1 mole of N$_2$ reacts with 3 moles of H$_2$.
For 2 moles of N$_2$, required H$_2$ = $3 \times 2 = 6$ moles.
Only 5 moles of H$_2$ are available, which is less than required (6 moles). Hence, H$_2$ is the limiting reagent.

Key Concept: Mole Concept and Avogadro's Number

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the mass of 1 mole of carbon-12 is exactly 12 grams, which is true because it is the definition of the mole in terms of carbon-12. The reason states that the atomic mass unit is defined as one-twelfth the mass of a carbon-12 atom, which is also true. However, the reason does not explain why the mass of 1 mole of carbon-12 is 12 grams; instead, it provides an independent definition. Hence, both are true, but the reason is not the correct explanation of the assertion.

Key Concept: Percentage Composition, Empirical Formula

c) C$_2$H$_2$O$_4$
[Solution Description]
Empirical formula calculation:
Assume 100 g of the compound:
Mass of carbon = 26.59 g, mass of hydrogen = 2.22 g, mass of oxygen = 71.19 g.
Moles of carbon = $\frac{26.59}{12} \approx 2.22$.
Moles of hydrogen = $\frac{2.22}{1} = 2.22$.
Moles of oxygen = $\frac{71.19}{16} \approx 4.45$.
Divide by smallest (2.22):
Carbon ratio = $\frac{2.22}{2.22} = 1$.
Hydrogen ratio = $\frac{2.22}{2.22} = 1$.
Oxygen ratio = $\frac{4.45}{2.22} \approx 2$.
Empirical formula = CHO$_2$, with empirical weight = $12 + 1 + 32 = 45$.
Calculate molecular weight:
Molecular weight = $2 \times \text{V.D.} = 2 \times 45 = 90$.
$n = \frac{90}{45} = 2$.
Molecular formula = (CHO$_2$)$_2$ = C$_2$H$_2$O$_4$.

Key Concept: Avogadro’s Law, Atomicity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) correctly states the volume ratio for the reaction $H_2 + Cl_2 \rightarrow 2HCl$ as 1:1:2, which is consistent with Gay-Lussac’s law. This ratio implies that one molecule of $H_2$ reacts with one molecule of $Cl_2$ to produce two molecules of $HCl$. Since two molecules of $HCl$ are formed, each reactant must consist of diatomic molecules to satisfy stoichiometry (i.e., $H_2$ and $Cl_2$).
The reason (R) correctly states Avogadro’s law, which explains why the volume ratio corresponds to the molecular ratio. By combining Avogadro’s law with the observed volume ratio, we can deduce the atomicity of the gases involved. Thus, (R) is the correct explanation for (A).

Key Concept: Percentage Composition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To calculate the percentage composition of oxygen in $H_2O$:
Molecular mass of $H_2O = 2 \times 1 + 16 = 18$ g/mol.
Mass of oxygen in one molecule = 16 g.
Percentage of oxygen = $\frac{16}{18} \times 100 = 88.89\%$.
The Assertion and Reason are both true, and the Reason correctly explains the Assertion because the calculation matches the given percentage.

Key Concept: Empirical Formula Calculation

c) $CH_2O$
[Solution Description]
Step 1: Assume the total mass of the compound is 100 g. So, we have 40 g of carbon, 6.7 g of hydrogen, and 53.3 g of oxygen.
Step 2: Calculate the number of moles of each element.
Moles of carbon = $\frac{40}{12} = 3.33$,
Moles of hydrogen = $\frac{6.7}{1} = 6.7$,
Moles of oxygen = $\frac{53.3}{16} = 3.33$.
Step 3: Divide each by the smallest number of moles to get the simplest ratio.
Carbon ratio = $\frac{3.33}{3.33} = 1$,
Hydrogen ratio = $\frac{6.7}{3.33} = 2$,
Oxygen ratio = $\frac{3.33}{3.33} = 1$.
Therefore, the empirical formula is $CH_2O$.

Key Concept: Drug Stability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that aspirin should be stored in a cool, dry place. This is correct because aspirin is susceptible to hydrolysis when exposed to moisture. The reason explains that hydrolysis breaks aspirin into salicylic acid and acetic acid, which diminishes its efficacy. Thus, both statements are true, and the reason correctly justifies the assertion.

Key Concept: Temperature Conversion

b) 298 K
[Solution Description]
To convert Celsius to Kelvin, use the formula $K = ^{\circ}C + 273$. For $25^{\circ}C$:
$K = 25 + 273 = 298 \, K$

Key Concept: Basic Pharmaceutical Term

b) A solid compressed dosage form
[Solution Description]
A tablet is a solid dosage form containing medicinal substances with or without suitable diluents. It is compressed into a solid form for oral administration.

Key Concept: Mole Concept, Gram Molecular Mass

b) 2 moles
[Solution Description]
First, calculate the molar mass of $H_2SO_4$:
$\text{Molar Mass} = 2 \times 1 \text{ g/mol (H)} + 32 \text{ g/mol (S)} + 4 \times 16 \text{ g/mol (O)} = 98 \text{ g/mol}$
Find the number of moles in 49 g:
$\text{Moles} = \frac{49 \text{ g}}{98 \text{ g/mol}} = 0.5 \text{ moles of } H_2SO_4$
Each $H_2SO_4$ molecule contains 4 O atoms. Thus, total moles of O atoms:
$0.5 \text{ moles} \times 4 = 2 \text{ moles of O atoms}$

Key Concept: Gram Molecular Mass

b) 28 g
[Solution Description]
Given:
280 cm³ of gas weighs 0.35 g.
22,400 cm³ (molar volume) would weigh:
$\frac{0.35 \times 22400}{280} = 28$ g.
Hence, the gram molecular mass is 28 g.

Key Concept: Empirical and Molecular Formula

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The empirical formula weight of $\text{CH}_2\text{O}$ is calculated as:
$\text{Carbon: } 1 \times 12 = 12$
$\text{Hydrogen: } 2 \times 1 = 2$
$\text{Oxygen: } 1 \times 16 = 16$
Total empirical formula weight $= 12 + 2 + 16 = 30$.
The value of $n$ is:
$n = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} = \frac{180}{30} = 6$
Thus, the molecular formula is:
$\text{CH}_2\text{O} \times 6 = \text{C}_6\text{H}_{12}\text{O}_6$
Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Gay-Lussac’s Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia is $N_2 + 3H_2 \rightarrow 2NH_3$. According to Gay-Lussac’s Law, the volumes of gases react in simple whole number ratios under the same temperature and pressure conditions. Here, 1 volume of nitrogen reacts with 3 volumes of hydrogen. Therefore, if 50 cm$^3$ of nitrogen is used, the volume of hydrogen required would be $50 \times 3 = 150$ cm$^3$, as stated in Assertion (A). Since the reason (R) correctly explains the assertion (A), both are true, and Reason is the correct explanation of Assertion.

Key Concept: Definition of Mole

b) $6.022 \times 10^{23}$
[Solution Description]
Avogadro’s number ($N_A$) is defined as $6.022 \times 10^{23}$ particles per mole. This is a fundamental constant in chemistry used to relate macroscopic and microscopic quantities.

Key Concept: Identifying Excess Reactant

a) $N_2$
[Solution Description]
To find the excess reactant:
1. The stoichiometric ratio of $N_2$ to $H_2$ is 1:3.
2. Given 2 moles of $N_2$, the required $H_2$ is $2 \times 3 = 6$ moles.
3. Since only 5 moles of $H_2$ are available, $H_2$ is the limiting reactant.
4. Therefore, $N_2$ remains in excess.

Key Concept: Volume Calculation at STP

b) 22.4 L
[Solution Description]
The molar mass of water ($H_2O$) is $(2 \times 1) + 16 = 18 \, \text{g/mol}$. Given 36 g of water, this corresponds to $36 / 18 = 2$ moles. From the balanced equation, 2 moles of water produce 1 mole of $O_2$. At STP, 1 mole of any gas occupies 22.4 L. Therefore, the volume of $O_2$ produced is 22.4 L.

Key Concept: Avogadro's Number, Molar Mass

b) 44 g/mol
[Solution Description]
First, we know that at STP, 1 mole of any gas occupies 22.4 liters. The given volume is 5.6 liters.
Step 1: Calculate number of moles:
$\text{moles} = \frac{\text{volume}}{\text{molar volume}} = \frac{5.6}{22.4} = 0.25 \text{ moles}$
Step 2: Calculate molar mass:
$\text{molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{11}{0.25} = 44 \text{ g/mol}$

Key Concept: Empirical Formula

c) $CH_2O$
[Solution Description]
Assume 100 g of the compound.
Mass of carbon = 40 g, moles of carbon = $\frac{40}{12} = 3.33$ mol.
Mass of hydrogen = 6.7 g, moles of hydrogen = $\frac{6.7}{1} = 6.7$ mol.
Mass of oxygen = 53.3 g, moles of oxygen = $\frac{53.3}{16} = 3.33$ mol.
Simplest ratio: C: $3.33/3.33 = 1$, H: $6.7/3.33 \approx 2$, O: $3.33/3.33 = 1$.
Empirical formula is $CH_2O$.

Key Concept: STP Conditions

b) $0^\circ \text{C}, 760 \text{mmHg}$
[Solution Description]
STP (Standard Temperature and Pressure) is defined as $0^\circ \text{C}$ ($273 \text{K}$) for temperature and $1 \text{atm}$ (or $760 \text{mmHg}$) for pressure.

Key Concept: Moles to Particles, Molar Mass

b) $1.2044 \times 10^{24}$ molecules
[Solution Description]
- First, calculate the molar mass of $O_2$. Each oxygen atom has a molar mass of 16 g/mol, so $O_2$ has a molar mass of $2 \times 16 = 32$ g/mol.
- Next, find the number of moles in 64 g of $O_2$ using the formula:
$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{64}{32} = 2 \text{ moles}$
- Finally, calculate the number of molecules using Avogadro’s number:
$\text{Number of Molecules} = 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}$

Key Concept: Mole Concept

b) 63 g
[Solution Description]
The molar mass of water ($H_2O$) is 18 g/mol. To find the mass of 3.5 moles, multiply the number of moles by the molar mass:
$\text{Mass} = \text{Number of moles} \times \text{Molar mass} = 3.5 \times 18 = 63 \text{ g}$

Key Concept: Calcium Carbide in Fruit Ripening

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Calcium carbide ($CaC_2$) reacts with moisture to form acetylene gas ($C_2H_2$), which mimics the natural plant hormone ethylene, thereby accelerating fruit ripening. The assertion is true because calcium carbide is indeed used for artificial ripening. The reason is also true as acetylene acts similarly to ethylene, a natural ripening hormone. However, the reason is not entirely correct because while acetylene mimics ethylene, it is not itself a plant hormone. Thus, the correct answer is option b).

Key Concept: Avogadro's Law

b) The same number of molecules
[Solution Description]
Avogadro’s law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules." This means the number of molecules is the same regardless of the gas type.

Key Concept: Gay-Lussac's Law of Combining Volumes, Volume Ratio Calculation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
We analyze the given chemical reaction and apply Gay-Lussac's law step by step:
The balanced equation is: $2CO + O_2 \rightarrow 2CO_2$
This shows that 2 volumes of CO react with 1 volume of O$_2$ to produce 2 volumes of CO$_2$.
Given volume of CO = 50 mL
From the stoichiometric ratio (2:1), volume of O$_2$ required = $\frac{1}{2}$ × volume of CO = $\frac{1}{2}$ × 50 mL = 25 mL
The reason correctly states Gay-Lussac's law about volume ratios in gaseous reactions. Moreover, it correctly explains why we get 25 mL oxygen for 50 mL CO.
Both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Mass from Molarity

b) 18 g
[Solution Description]
First, calculate the moles of glucose required using the molarity formula rearranged as $\text{Moles} = M \times V$, where $V$ is in liters. Here, $V = 200 \, \text{mL} = 0.2 \, \text{L}$, so $\text{Moles} = 0.5 \, \text{M} \times 0.2 \, \text{L} = 0.1 \, \text{mol}$. Next, convert moles to grams: $\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.1 \, \text{mol} \times 180 \, \text{g/mol} = 18 \, \text{g}$.

Key Concept: Molar Mass and Mole Concept

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The molar mass of a substance is numerically equal to its molecular mass expressed in grams per mole. The molecular mass of water ($H_2O$) is calculated as $2 \times 1$ (for hydrogen) + $16$ (for oxygen) = 18 a.m.u. Therefore, the molar mass of water is 18 g/mol. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Mass-Mass Relationship, Percentage Yield

c) 78.57\%
[Solution Description]
First find theoretical yield:
- Moles of $Fe_2O_3$ = $\frac{500000}{160} = 3125$ mol
- Moles of C = $\frac{100000}{12} \approx 8333.33$ mol
Check limiting reagent:
- $C$ required for 3125 mol $Fe_2O_3$ = $\frac{3}{2} \times 3125 = 4687.5$ mol
- Since available C > required, $Fe_2O_3$ is limiting
Theoretical yield of Fe:
- Moles of Fe = $2 \times 3125 = 6250$ mol
- Mass of Fe = $6250 \times 56 = 350000$ g = 350 kg
Calculate percentage yield:
- $\frac{275}{350} \times 100 = 78.57\%$

Key Concept: STP, Gram Molecular Mass

c) 28 g
[Solution Description]
500 cm$^3$ of the gas at STP weighs 0.625 g.
Therefore, 22400 cm$^3$ (molar volume at STP) will weigh:
$\text{Gram Molecular Mass} = \frac{0.625 \times 22400}{500} = 28 \text{ g}$

Key Concept: Molarity Calculation

c) 0.25 M
[Solution Description]
Molarity ($M$) is calculated using the formula:
$M = \frac{\text{moles of solute}}{\text{liters of solution}}$
Given: moles of solute = 0.5, liters of solution = 2
So, $M = \frac{0.5}{2} = 0.25$ M

Key Concept: Empirical Formula, Molecular Weight

d) C$_6$H$_{12}$O$_6$
[Solution Description]
First, find the empirical formula:
Assume 100 g of the compound:
Mass of carbon = 40 g, mass of hydrogen = 6.7 g, mass of oxygen = 53.3 g.
Moles of carbon = $\frac{40}{12} \approx 3.33$.
Moles of hydrogen = $\frac{6.7}{1} = 6.7$.
Moles of oxygen = $\frac{53.3}{16} \approx 3.33$.
Divide by the smallest number of moles (3.33):
Carbon ratio = $\frac{3.33}{3.33} = 1$.
Hydrogen ratio = $\frac{6.7}{3.33} \approx 2$.
Oxygen ratio = $\frac{3.33}{3.33} = 1$.
Empirical formula = CH$_2$O, with empirical weight = $12 + 2 + 16 = 30$.
Find $n$ using molecular weight:
$n = \frac{180}{30} = 6$.
Molecular formula = (CH$_2$O)$_6$ = C$_6$H$_{12}$O$_6$.

Key Concept: Balancing equations, stoichiometry, mole concept

d) 44.8 L
[Solution Description]
First, calculate the moles of $NH_3$:
Molar mass of $NH_3 = 14 + 3 \times 1 = 17\,g/mol$.
Moles of $NH_3 = \frac{34\,g}{17\,g/mol} = 2\,mol$.
From the balanced equation, 4 moles of $NH_3$ produce 4 moles of $NO$.
Thus, 2 moles of $NH_3$ will produce 2 moles of $NO$.
Volume of 1 mole of gas at STP = $22.4\,L$.
Volume of 2 moles of $NO$ = $2 \times 22.4\,L = 44.8\,L$.

Key Concept: Charles's Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that doubling the temperature from $273\,K$ to $546\,K$ doubles the volume of a gas at constant pressure. This aligns with Charles’s Law, which mathematically states $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ at constant pressure. Here, $T_2 = 2T_1$ ($546\,K = 2 \times 273\,K$), so $V_2 = 2V_1$. Thus, the assertion is true. The reason correctly explains Charles’s Law and justifies the assertion, making it the correct explanation.

Key Concept: Balancing Chemical Equations

a) 1, 5, 3, 4
[Solution Description]
To balance the equation, start with carbon atoms. There are 3 carbons on the left, so place a 3 before $CO_2$. Next, balance hydrogen atoms: there are 8 hydrogens on the left, so place a 4 before $H_2O$. Now, count the oxygen atoms on the right: $(3 \times 2) + (4 \times 1) = 10$ oxygens. Therefore, place a 5 before $O_2$ to balance it. The balanced equation is $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$.

Key Concept: Molar Volume, Stoichiometry in Gas Reactions

d) 44.8 dm$^3$
[Solution Description]
The balanced equation is:
$CaCO_3 \rightarrow CaO + CO_2$
1 mole of $CaCO_3$ produces 1 mole of $CO_2$.
Moles of $CaCO_3 = \frac{200 \text{ g}}{100 \text{ g/mol}} = 2$ moles.
At STP, 1 mole of gas occupies 22.4 dm$^3$.
Volume of $CO_2$ produced = $2 \times 22.4 = 44.8$ dm$^3$.

Key Concept: Z-score

c) It is 2 standard deviations above the mean
[Solution Description]
A z-score of 2 means the value is 2 standard deviations above the mean. For a normal distribution, this corresponds to ~97.7th percentile.

Key Concept: Mass from Moles

c) 36 g
[Solution Description]
The formula to calculate mass from moles is:
$\text{Mass (in grams)} = \text{Moles} \times \text{Molecular Mass}$
Given, Moles = 2, Molecular Mass of $H_2O$ = 18 u.
$\text{Mass} = 2 \times 18 = 36 \text{ g}$

Key Concept: Relative Vapour Density

c) 32
[Solution Description]
Relative molecular mass is calculated as $2 \times$ Relative Vapour Density (V.D.). Given V.D. = 16, so,
$\text{Rel. Molecular Mass} = 2 \times 16 = 32$.
The correct answer is 32.

Key Concept: Drug Administration Route

d) Directly into a vein
[Solution Description]
Intravenous (IV) medications are directly injected into a vein, allowing rapid absorption into the bloodstream.

Key Concept: Normality, Redox Reactions

c) 0.5 N
[Solution Description]
In acidic medium, MnO$_4^-$ + 8H$^+$ + 5e$^-$ → Mn$^{2+}$ + 4H$_2$O. Here, n-factor for KMnO$_4$ = 5.
Molarity = 0.1 M ⇒ Normality = Molarity × n-factor = 0.1 × 5 = 0.5 N.