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Tuesday, 31 March 2026

NCERT Solutions: Chapter 1: Some Basic Concepts of Chemistry.

Molarity Calculation: 4g NaOH in 250mL Solution | Edunes Online Education

  Edunes Online Education

1.1 Calculate the molar mass of the following:

  • (i) $H_2O$: $(2 \times 1.008) + 16.00 = \mathbf{18.016 \, \text{g/mol}}$

  • (ii) $CO_2$: $12.01 + (2 \times 16.00) = \mathbf{44.01 \, \text{g/mol}}$

  • (iii) $CH_4$: $12.01 + (4 \times 1.008) = \mathbf{16.042 \, \text{g/mol}}$


1.2 Calculate the mass per cent of different elements present in sodium sulphate ($Na_2SO_4$).

Step 1: Find Molar Mass of $Na_2SO_4$

$(2 \times 23.0) + 32.06 + (4 \times 16.00) = 142.06 \, \text{g/mol}$

Step 2: Calculate Mass % ($\dfrac{\text{Mass of element}}{\text{Molar mass}} \times 100$)

  • Sodium (Na): $\dfrac{46.0}{142.06} \times 100 = \mathbf{32.38\%}$

  • Sulphur (S): $\dfrac{32.06}{142.06} \times 100 = \mathbf{22.57\%}$

  • Oxygen (O): $\dfrac{64.0}{142.06} \times 100 = \mathbf{45.05\%}$


1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Element Fe O
Mass % 69.9 30.1
Atomic Mass 55.85 16.00
Moles
(Mass/At. Mass) 		\( \dfrac{69.9}{55.85} = 1.25 \) \( \dfrac{30.1}{16.00} = 1.88 \)
Molar Ratio \( \dfrac{1.25}{1.25} = 1 \) \( \dfrac{1.88}{1.25} = 1.5 \)
Simple Ratio 2 3

Empirical Formula: $\mathbf{Fe_2O_3}$


1.4 Calculate the amount of carbon dioxide produced when:

Reaction: $C(s) + O_2(g) \rightarrow CO_2(g)$

  • (i) 1 mole of carbon is burnt in air: Since $O_2$ is in excess, 1 mole of $C$ gives 1 mole (44g) of $CO_2$.

  • (ii) 1 mole of carbon is burnt in 16g of dioxygen: 16g $O_2 = 0.5$ mole. $O_2$ is the limiting reagent. 0.5 mole $O_2$ produces 0.5 mole (22g) of $CO_2$.

  • (iii) 2 moles of carbon are burnt in 16g of dioxygen: Again, 16g $O_2 (0.5 \text{ mole})$ is the limiting reagent. It can only react with 0.5 mole of $C$ to produce 0.5 mole (22g) of $CO_2$.


1.5 Calculate the mass of sodium acetate ($CH_3COONa$) required to make 500 mL of 0.375 molar aqueous solution.

Step 1: Use Molarity formula ($M = \dfrac{n}{V}$)

$0.375 = \dfrac{n}{0.500 \, \text{L}} \implies n = 0.375 \times 0.500 = 0.1875 \, \text{moles}$

Step 2: Convert moles to mass (Molar mass = 82.0245 g/mol)

$\text{Mass} = 0.1875 \, \text{mol} \times 82.0245 \, \text{g/mol} = \mathbf{15.38 \, \text{g}}$


1.6 Calculate the concentration of nitric acid ($HNO_3$) in moles per litre ($M$) in a sample with density 1.41 g/mL and 69% mass per cent.

Step 1: Find mass of 1L solution

$\text{Mass} = \text{Volume} \times \text{Density} = 1000 \, \text{mL} \times 1.41 \, \text{g/mL} = 1410 \, \text{g}$

Step 2: Find mass of $HNO_3$ in that 1L

$69\% \text{ of } 1410 \, \text{g} = 0.69 \times 1410 = 972.9 \, \text{g}$

Step 3: Calculate Molarity (Molar mass of $HNO_3 = 63 \, \text{g/mol}$)

$M = \dfrac{972.9 \, \text{g} / 63 \, \text{g/mol}}{1 \, \text{L}} = \mathbf{15.44 \, \text{M}}$



1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL⁻¹ and the mass per cent of nitric acid in it being 69%.

Step 1: Find the mass of 1 L of the solution

  • $\text{Density} = 1.41\text{ g/mL}$

  • $\text{Mass of 1000 mL solution} = 1000 \times 1.41 = 1410\text{ g}$

Step 2: Find the mass of actual } HNO_3 \text{ in the solution

  • The solution is $69\%$ $HNO_3$ by mass.

  • $\text{Mass of } HNO_3 = \dfrac{69}{100} \times 1410\text{ g} = 972.9\text{ g}$

Step 3: Calculate Molarity (Moles per litre)

  • $\text{Molar mass of } HNO_3 = 1 + 14 + (3 \times 16) = 63\text{ g/mol}$

  • $\text{Moles of } HNO_3 = \dfrac{972.9\text{ g}}{63\text{ g/mol}} = 15.44\text{ mol}$

  • Concentration = 15.44 M

1.7 How much copper can be obtained from 100 g of copper sulphate ($CuSO_4$)?

Step 1: Calculate the molar mass of $CuSO_4$

  • $\text{Atomic mass of Cu} = 63.5\text{ u}$, $\text{S} = 32\text{ u}$, $\text{O} = 16\text{ u}$

  • $\text{Molar mass} = 63.5 + 32 + (4 \times 16) = 159.5\text{ g/mol}$

Step 2: Use the ratio to find the mass of Copper

  • $159.5\text{ g}$ of $CuSO_4$ contains $63.5\text{ g}$ of Copper.

  • $\text{Copper in 100 g } CuSO_4 = \left(\dfrac{63.5}{159.5}\right) \times 100 = \mathbf{39.81\text{ g}}$

1.8 Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

(Note: Molar mass of this oxide is 159.69 g/mol)

Step 1: Find the Empirical Formula

  • $\text{Moles of Fe} = \dfrac{69.9}{55.85} = 1.25$

  • $\text{Moles of O} = \dfrac{30.1}{16.00} = 1.88$

  • $\text{Simple molar ratio (Fe:O)} = 1 : 1.5$, which simplifies to 2 : 3.

  • $\text{Empirical Formula} = Fe_2O_3$

Step 2: Determine Molecular Formula

  • $\text{Empirical Formula Mass} = (2 \times 55.85) + (3 \times 16) = 159.7\text{ g/mol}$

  • Since the molar mass ($159.69$) is approximately equal to the empirical mass, $n = 1$.

  • Molecular Formula = $Fe_2O_3$

1.9 Calculate the atomic mass (average) of chlorine using the following data:

Isotope % Natural Abundance Molar Mass (g/mol)
$^{35}Cl$ 75.77 34.9689
$^{37}Cl$ 24.23 36.9659

Calculation:

$\text{Avg. Mass}$

$= \dfrac{(75.77 \times 34.9689) + (24.23 \times 36.9659)}{100}$
$\text{Avg. Mass}$

$= \dfrac{2649.59 + 895.68}{100} = \dfrac{3545.27}{100}$

$= \mathbf{35.4527\text{ u}}$

1.10 In three moles of ethane ($C_2H_6$), calculate the following:

  • (i) Number of moles of carbon atoms:

    1 mole of $C_2H_6$ has 2 moles of C.

    3 moles of $C_2H_6$ have $3 \times 2 = \mathbf{6\text{ moles}}$ of C atoms.

  • (ii) Number of moles of hydrogen atoms:

    1 mole of $C_2H_6$ has 6 moles of H.

    3 moles of $C_2H_6$ have $3 \times 6 = \mathbf{18\text{ moles}}$ of H atoms.

  • (iii) Number of molecules of ethane:

    $1\text{ mole} = 6.022 \times 10^{23}\text{ molecules}$

    $3\text{ moles} = 3 \times 6.022 \times 10^{23}$

    $= \mathbf{1.806 \times 10^{24}\text{ molecules}}$

To solve these chemistry problems from your NCERT textbook , we will use a systematic problem-solving model: Identify (what is given), Strategize (the formula needed), and Execute (the calculation).

1.11 Concentration of Sugar

Problem: What is the concentration of sugar ($C_{12}H_{22}O_{11}$) in $mol \ L^{-1}$ if $20 \ g$ are dissolved in enough water to make a final volume up to $2 \ L$?

  • Identify:

    • Mass of sugar = $20 \ g$

    • Volume of solution = $2 \ L$

    • Molar mass of $C_{12}H_{22}O_{11}$ $ = (12 \times 12) + (22 \times 1) + (11 \times 16)$ $ = 342 \ g \ mol^{-1}$

  • Strategize: Molarity ($M$) = $ \dfrac{\text{moles of solute}}{\text{Volume of solution in Litres}}$

  • Execute:

    1. Calculate moles: $n = \dfrac{20 \ g}{342 \ g \ mol^{-1}} \approx 0.0585 \ mol$

    2. Calculate Molarity: $M = \dfrac{0.0585 \ mol}{2 \ L} = 0.02925 \ mol \ L^{-1}$

1.12 Volume of Methanol Needed

Problem: If the density of methanol is $0.793 \ kg \ L^{-1}$, what is its volume needed for making $2.5 \ L$ of its $0.25 \ M$ solution?

  • Identify:

    • Target Molarity ($M_2$) = $0.25 \ M$

    • Target Volume ($V_2$) = $2.5 \ L$

    • Density of Methanol ($CH_3OH$) = $0.793 \ kg \ L^{-1} = 793 \ g \ L^{-1}$

    • Molar mass of $CH_3OH = 12 + (4 \times 1) + 16 = 32 \ g \ mol^{-1}$

  • Strategize: 1. Find the molarity of pure methanol ($M_1$) using density.

    2. Use the dilution formula: $M_1V_1 = M_2V_2$.

  • Execute:

    1. $M_1 = \dfrac{\text{Density}}{\text{Molar Mass}} = \dfrac{793 \ g \ L^{-1}}{32 \ g \ mol^{-1}} = 24.78 \ M$

    2. $24.78 \times V_1 = 0.25 \times 2.5$

    3. $V_1 = \dfrac{0.625}{24.78} \approx 0.0252 \ L = 25.2 \ mL$

1.13 Pressure at Sea Level

Problem: If mass of air at sea level is $1034 \ g \ cm^{-2}$, calculate the pressure in pascal ($Pa$).

  • Identify:

    • Mass ($m$) = $1034 \ g = 1.034 \ kg$

    • Area ($A$) = $1 \ cm^2 = 10^{-4} \ m^2$

    • $g$ (acceleration due to gravity) $\approx 9.8 \ m \ s^{-2}$

  • Strategize: Pressure ($P$) = $\dfrac{\text{Force}}{\text{Area}} = \dfrac{m \times g}{A}$

  • Execute:

    1. $P = \dfrac{1.034 \ kg \times 9.8 \ m \ s^{-2}}{10^{-4} \ m^2}$

    2. $P = 1.01332 \times 10^5 \ N \ m^{-2} = 1.01332 \times 10^5 \ Pa$

1.14 SI Unit of Mass

Problem: What is the SI unit of mass? How is it defined?

  • Answer: * The SI unit of mass is the kilogram (kg).

    • Definition: It is defined by taking the fixed numerical value of the Planck constant $h$ to be $6.62607015 \times 10^{-34}$ when expressed in the unit $J \ s$, which is equal to $kg \ m^2 \ s^{-1}$.

1.15 Match Prefixes with Multiples

Problem: Match the following prefixes with their multiples.

Prefix Multiple Correct Match
(i) micro $10^6$ $10^{-6}$
(ii) deca $10^9$ $10^1$
(iii) mega $10^{-6}$ $10^6$
(iv) giga $10^{-15}$ $10^9$
(v) femto $10^1$ $10^{-15}$

Quick Tip: When dealing with molarity and density (like in 1.12), always ensure your units for density ($g/L$ or $g/mL$) match the volume units you are using in your molarity calculation!

1.16 Significant Figures

Problem: What do you mean by significant figures?

  • Identify: Concept of precision in measurements.

  • Strategize: Define based on scientific measurement standards.

  • Execute: Significant figures are the total number of digits in a value, typically a measurement, that contribute to its degree of accuracy. This includes all certain digits plus one final digit that is somewhat uncertain or estimated.

1.17 Chloroform Contamination

Problem: A sample of drinking water was contaminated with chloroform ($CHCl_3$) at $15 \ ppm$ (by mass).

(i) Express this in percent by mass.

(ii) Determine the molality of chloroform in the water sample.

  • Identify: * Concentration = $15 \ ppm$ (parts per million).

    • Molar mass of $CHCl_3 = 12 + 1 + (3 \times 35.5) = 119.5 \ g \ mol^{-1}$.

  • Strategize: * $ppm$ means grams of solute per $10^6 \ g$ of solution.

    • Percent by mass = $\dfrac{\text{Mass of solute}}{\text{Total mass}} \times 100$.

    • Molality ($m$) = $\dfrac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$.

  • Execute:

    1. Mass Percent: $\dfrac{15}{10^6} \times 100 = 1.5 \times 10^{-3} \%$.

    2. Molality: * Moles of $CHCl_3 = \dfrac{15 \ g}{119.5 \ g \ mol^{-1}} \approx 0.125 \ mol$.

      • Mass of solvent $\approx 10^6 \ g = 1000 \ kg$ (since $15 \ g$ is negligible).

      • $m = \dfrac{0.125 \ mol}{1000 \ kg} = 1.25 \times 10^{-4} \ m$.

1.18 Scientific Notation

Problem: Express the following in scientific notation ($N \times 10^n$):

  • Strategize: Move the decimal to follow the first non-zero digit.

  • Execute:

    • (i) 0.0048 $\rightarrow 4.8 \times 10^{-3}$

    • (ii) 234,000 $\rightarrow 2.34 \times 10^5$

    • (iii) 8008 $\rightarrow 8.008 \times 10^3$

    • (iv) 500.0 $\rightarrow 5.000 \times 10^2$

    • (v) 6.0012 $\rightarrow 6.0012 \times 10^0$

1.19 Counting Significant Figures

Problem: How many significant figures are present in the following?

  • Strategize: Apply rules (leading zeros don't count; trailing zeros after decimals do; zeros between digits do).

  • Execute:

    • (i) 0.0025: 2 (2, 5)

    • (ii) 208: 3 (2, 0, 8)

    • (iii) 5005: 4 (all digits)

    • (iv) 126,000: 3 (trailing zeros without a decimal are usually non-significant)

    • (v) 500.0: 4 (trailing zero after decimal is significant)

    • (vi) 2.0034: 5 (all digits)

1.20 Rounding Off

Problem: Round up the following up to three significant figures:

  • Strategize: Look at the 4th digit; if $\geq 5$, round up.

  • Execute:

    • (i) 34.216: $34.2$ (4th digit is 1)

    • (ii) 10.4107: $10.4$ (4th digit is 1)

    • (iii) 0.04597: $0.0460$ (4th digit is 7, rounds 9 up)

    • (iv) 2808: $2810$ or $2.81 \times 10^3$ (4th digit is 8)

Key Insight: In chemistry, significant figures aren't just "math rules"—they represent the actual precision of the tools used in the lab. Always check if a zero is a "placeholder" or a "measured value"!


Question 1.21

Data provided for dinitrogen and dioxygen compounds:

  • (i) 14 g $N_2$ + 16 g $O_2$

  • (ii) 14 g $N_2$ + 32 g $O_2$

  • (iii) 28 g $N_2$ + 32 g $O_2$

  • (iv) 28 g $N_2$ + 80 g $O_2$

(a) Which law of chemical combination is obeyed?

  1. Fix the mass of one element: Let's fix dinitrogen at 28 g.

  2. Adjust the masses:

    • (i) 28 g $N_2$ would react with 32 g $O_2$

    • (ii) 28 g $N_2$ would react with 64 g $O_2$

    • (iii) 28 g $N_2$ reacts with 32 g $O_2$

    • (iv) 28 g $N_2$ reacts with 80 g $O_2$

  3. Ratio of Oxygen: The masses of oxygen combining with a fixed mass of nitrogen are 32, 64, 32, and 80. The ratio is $32:64:80$, which simplifies to $2:4:5$ (a simple whole-number ratio).

  4. Conclusion: This obeys the Law of Multiple Proportions.

    Statement: If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

(b) Conversions:

  • (i) 1 km = $10^6$ mm = $10^{12}$ pm

  • (ii) 1 mg = $10^{-6}$ kg = $10^6$ ng

  • (iii) 1 mL = $10^{-3}$ L = $10^{-3}$ $dm^3$


Question 1.22

Calculate the distance covered by light in 2.00 ns if speed is $3.0 \times 10^8$ m/s.

  1. Identify values:

    • Speed ($v$) = $3.0 \times 10^8$ m/s

    • Time ($t$) = 2.00 ns = $2.00 \times 10^{-9}$ s

  2. Formula: $Distance = Speed \times Time$

  3. Calculation:

    $$Distance = (3.0 \times 10^8 \, \text{m/s}) \times (2.00 \times 10^{-9} \, \text{s})$$
    $$Distance = 6.00 \times 10^{-1} \, \text{m} = \mathbf{0.600 \, \text{m}}$$


Question 1.23

Identify the limiting reagent in $A + B_2 \rightarrow AB_2$:

  • (i) 300 atoms of A + 200 molecules of B: 1 atom of A needs 1 molecule of B. Here, B is less. Limiting Reagent: B.

  • (ii) 2 mol A + 3 mol B: 1 mol A needs 1 mol B. Here, A is less. Limiting Reagent: A.

  • (iii) 100 atoms of A + 100 molecules of B: Both are in the exact stoichiometric ratio. No limiting reagent.

  • (iv) 5 mol A + 2.5 mol B: B is half of what is needed for A. Limiting Reagent: B.

  • (v) 2.5 mol A + 5 mol B: A is half of what is needed for B. Limiting Reagent: A.


Question 1.24

$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

(Note: The equation in the snippet was $N_2 + H_2 \rightarrow 2NH_3$, but the balanced form is $N_2 + 3H_2$)

(i) Mass of $NH_3$ produced from $2.00 \times 10^3$ g $N_2$ and $1.00 \times 10^3$ g $H_2$:

  1. Moles of $N_2$: $2000 / 28 = 71.43$ mol

  2. Moles of $H_2$: $1000 / 2.016 = 496.03$ mol

  3. Limiting Reagent: 71.43 mol $N_2$ requires $71.43 \times 3 = 214.29$ mol $H_2$. Since we have 496.03 mol, $N_2$ is the limiting reagent.

  4. $NH_3$ Produced: $71.43 \text{ mol } N_2 \times 2 = 142.86$ mol $NH_3$.

    • $Mass = 142.86 \times 17.03 = \mathbf{2433 \, \text{g}}$ (approx).

(ii) Will any reactant remain?

Yes, Dihydrogen ($H_2$) will remain unreacted.

(iii) Mass of unreacted $H_2$:

  1. Used $H_2$: 214.29 mol.

  2. Remaining moles: $496.03 - 214.29 = 281.74$ mol.

  3. Remaining mass: $281.74 \times 2.016 = \mathbf{568 \, \text{g}}$.


Question 1.25

Difference between 0.50 mol $Na_2CO_3$ and 0.50 M $Na_2CO_3$:

  • 0.50 mol $Na_2CO_3$: This represents a specific amount/mass of the substance.

    • Molar mass of $Na_2CO_3 \approx 106$ g/mol.

    • Mass = $0.50 \times 106 = \mathbf{53 \, \text{g}}$.

  • 0.50 M $Na_2CO_3$: This represents molarity (concentration). It means 0.50 moles of sodium carbonate are dissolved in 1 Litre of solution.


Question 1.26

If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

  1. Write the balanced equation:

    $$2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$$

  2. Apply Gay Lussac’s Law: The volumes of reactants and products bear a simple whole-number ratio.

    • According to the equation: 2 volumes of $H_2$ react with 1 volume of $O_2$ to produce 2 volumes of $H_2O$.

  3. Calculate:

    • Given: 10 volumes of $H_2$ and 5 volumes of $O_2$.

    • Since the ratio is $2:1$, all 10 volumes of $H_2$ will react with exactly 5 volumes of $O_2$.

  4. Result: The volume of water vapour produced will be equal to the volume of $H_2$ consumed.

    • Volume of water vapour = 10 volumes.


Question 1.27

Convert the following into basic units (SI units):

  • (i) 28.7 pm:

    • $1 \text{ pm} = 10^{-12} \text{ m}$

    • $28.7 \times 10^{-12} \text{ m} = \mathbf{2.87 \times 10^{-11} \text{ m}}$

  • (ii) 15.15 pm:

    • $15.15 \times 10^{-12} \text{ m} = \mathbf{1.515 \times 10^{-11} \text{ m}}$

  • (iii) 25365 mg:

    • Basic unit of mass is kg. $1 \text{ mg} = 10^{-6} \text{ kg}$

    • $25365 \times 10^{-6} \text{ kg} = \mathbf{2.5365 \times 10^{-2} \text{ kg}}$


Question 1.28

Which one of the following will have the largest number of atoms?

  1. Formula: $\text{Number of atoms} = \frac{\text{Given mass}}{\text{Molar mass}} \times N_A$ (multiply by atomicity if it's a molecule).

  2. Calculations:

    • (i) 1 g Au (s): $1/197 \text{ mol} \approx 0.005 \text{ mol}$

    • (ii) 1 g Na (s): $1/23 \text{ mol} \approx 0.043 \text{ mol}$

    • (iii) 1 g Li (s): $1/7 \text{ mol} \approx \mathbf{0.142 \text{ mol}}$

    • (iv) 1 g of $Cl_2$ (g): $(1/71) \times 2 \text{ atoms} \approx 0.028 \text{ mol of atoms}$

  3. Conclusion: 1 g of Li (s) has the largest number of atoms because it has the smallest molar mass.


Question 1.29

Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

  1. Identify definitions:

    • Mole fraction of ethanol $(\chi_{eth}) = 0.040$

    • Mole fraction of water $(\chi_{H_2O}) = 1 - 0.040 = 0.960$

  2. Assume total moles = 1:

    • $n_{ethanol} = 0.040 \text{ mol}$

    • $n_{water} = 0.960 \text{ mol}$

  3. Find volume of solution: Since the solution is dilute and density of water is $1 \text{ g/mL}$:

    • Mass of water = $0.960 \text{ mol} \times 18 \text{ g/mol} = 17.28 \text{ g}$

    • Volume of water $\approx$ Volume of solution = $17.28 \text{ mL} = 0.01728 \text{ L}$

  4. Molarity ($M$):

    $$M = \frac{n_{ethanol}}{\text{Volume in Litres}} = \frac{0.040}{0.01728} \approx \mathbf{2.31 \text{ M}}$$


Question 1.30

What will be the mass of one $^{12}C$ atom in g?

  1. Known values:

    • Molar mass of $^{12}C = 12 \text{ g/mol}$

    • Avogadro’s number ($N_A$) = $6.022 \times 10^{23} \text{ atoms/mol}$

  2. Calculation:

    $$\text{Mass of 1 atom} = \frac{\text{Molar mass}}{N_A}$$
    $$\text{Mass} = \frac{12 \text{ g}}{6.022 \times 10^{23}} \approx \mathbf{1.99 \times 10^{-23} \text{ g}}$$


Question 1.31

How many significant figures should be present in the answer of the following calculations?

  • (i) $\frac{0.02856 \times 298.15 \times 0.112}{0.5785}$: In multiplication and division, the result should have the same number of significant figures as the term with the least significant figures. Here, 0.112 has 3 significant figures.

    • Answer: 3

  • (ii) $5 \times 5.364$: Here, 5 is an exact number (counting number). The precision is determined by 5.364, which has 4 significant figures.

    • Answer: 4

  • (iii) $0.0125 + 0.7864 + 0.0215$: In addition, the result should have the same number of decimal places as the term with the least decimal places. All terms have 4 decimal places.

    • Answer: 4 (The result is 0.8204)


Question 1.32

Calculate the molar mass of naturally occurring argon isotopes:

Isotope Isotopic molar mass Abundance
$^{36}Ar$ 35.96755 g/mol 0.337%
$^{38}Ar$ 37.96272 g/mol 0.063%
$^{40}Ar$ 39.9624 g/mol 99.600%

  1. Formula: $Avg. Molar Mass = \sum (\text{Molar Mass} \times \text{Abundance})$

  2. Calculation:

    $$M = \frac{(35.96755 \times 0.337) + (37.96272 \times 0.063) + (39.9624 \times 99.600)}{100}$$
    $$M = \frac{12.121 + 2.391 + 3980.255}{100} = \frac{3994.767}{100}$$

    • Average Molar Mass = 39.948 g/mol


Question 1.33

Calculate the number of atoms in:

  • (i) 52 moles of Ar: * $1 \text{ mole} = 6.022 \times 10^{23} \text{ atoms}$

    • $52 \times 6.022 \times 10^{23} = \mathbf{3.131 \times 10^{25} \text{ atoms}}$

  • (ii) 52 u of He:

    • Atomic mass of He = 4 u.

    • $\text{Number of atoms} = 52 / 4 = \mathbf{13 \text{ atoms}}$

  • (iii) 52 g of He:

    • Moles of He = $52 / 4 = 13 \text{ mol}$

    • $\text{Number of atoms} = 13 \times 6.022 \times 10^{23} = \mathbf{7.828 \times 10^{24} \text{ atoms}}$


Question 1.34

Welding gas analysis:

  1. Find mass of C and H:

    • Mass of C in 3.38 g $CO_2$ = $3.38 \times (12/44) = 0.9218 \text{ g}$

    • Mass of H in 0.690 g $H_2O$ = $0.690 \times (2.016/18) = 0.0772 \text{ g}$

  2. Empirical Formula:

    • Moles C = $0.9218 / 12 = 0.0768$

    • Moles H = $0.0772 / 1 = 0.0772$

    • Ratio C:H is $1:1$. Empirical Formula: CH

  3. Molar Mass:

    • 10.0 L weighs 11.6 g at STP.

    • 22.4 L (1 mole) weighs $(11.6 / 10.0) \times 22.4 = \mathbf{25.98 \text{ g/mol} \approx 26 \text{ g/mol}}$

  4. Molecular Formula:

    • Empirical mass (CH) = 13.

    • $n = 26 / 13 = 2$.

    • Molecular Formula: $C_2H_2$ (Ethyne)


Question 1.35

$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$

Mass of $CaCO_3$ for 25 mL of 0.75 M HCl?

  1. Moles of HCl:

    • $Moles = Molarity \times Volume(L) = 0.75 \times 0.025 = 0.01875 \text{ mol}$

  2. Stoichiometry: 2 moles HCl react with 1 mole $CaCO_3$.

    • Moles $CaCO_3$ needed = $0.01875 / 2 = 0.009375 \text{ mol}$

  3. Mass of $CaCO_3$:

    • $Mass = 0.009375 \times 100 \text{ g/mol} = \mathbf{0.9375 \text{ g}}$


Question 1.36

$4HCl(aq) + MnO_2(s) \rightarrow 2H_2O(l) + MnCl_2(aq) + Cl_2(g)$

Grams of HCl for 5.0 g $MnO_2$?

  1. Moles of $MnO_2$:

    • Molar mass $MnO_2 = 55 + (2 \times 16) = 87 \text{ g/mol}$

    • $Moles = 5.0 / 87 = 0.0575 \text{ mol}$

  2. Stoichiometry: 1 mole $MnO_2$ reacts with 4 moles HCl.

    • Moles HCl needed = $0.0575 \times 4 = 0.230 \text{ mol}$

  3. Mass of HCl:

    • $Mass = 0.230 \times 36.5 \text{ g/mol} = \mathbf{8.40 \text{ g}}$

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