Chapter 1 Solutions – NCERT Class 12 Chemistry

Question 1.1

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Definition: A solution is a homogeneous mixture of two or more substances whose composition can be varied within certain limits.

Types: Based on the physical state of the solvent, there are 9 types of solutions (3 Gaseous, 3 Liquid, 3 Solid).

Gaseous Solutions:
- Gas in Gas (Air)
- Liquid in Gas (Chloroform with N2)
- Solid in Gas (Camphor in N2)
Liquid Solutions:
- Gas in Liquid (Oxygen in water)
- Liquid in Liquid (Ethanol in water)
- Solid in Liquid (Glucose in water)
Solid Solutions:
- Gas in Solid (H2 in Palladium)
- Liquid in Solid (Amalgam of Hg with Na)
- Solid in Solid (Alloys like Brass)

Question 1.2

Give an example of a solid solution in which the solute is a gas.

Solution of Hydrogen gas in Palladium (or Platinum).

Question 1.3

Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.

(i) Mole Fraction ($x$): Ratio of the number of moles of a component to the total number of moles of all components in the solution. $x_A = n_A / (n_A + n_B)$.

(ii) Molality ($m$): Number of moles of solute per kilogram of the solvent. $m = n_{solute} / W_{solvent (kg)}$.

(iii) Molarity ($M$): Number of moles of solute dissolved in one litre of solution. $M = n_{solute} / V_{solution (L)}$.

(iv) Mass Percentage (w/w): Mass of the component per 100g of the solution. $\% = (w_{solute} / w_{total}) \times 100$.

Question 1.4

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g/mL?

Given

Mass % = 68% (68g $HNO_3$ in 100g solution).
Density $d = 1.504 \text{ g/mL}$.
Molar Mass of $HNO_3 = 1 + 14 + 48 = 63 \text{ g/mol}$.

Calculation

1. Moles of solute ($n$):

$$n = \frac{68}{63} \approx 1.079 \text{ mol}$$

2. Volume of solution ($V$):

$$V = \frac{\text{Mass}}{\text{Density}} = \frac{100}{1.504} \approx 66.49 \text{ mL} = 0.06649 \text{ L}$$

3. Molarity ($M$):

$$M = \frac{n}{V} = \frac{1.079}{0.06649} \approx 16.23 \text{ M}$$

Molarity = 16.23 M

Question 1.5

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g/mL, then what shall be the molarity of the solution?

1. Basic Data

10% w/w means 10g Glucose ($C_6H_{12}O_6$) in 90g Water.
Molar Mass Glucose = 180 g/mol. Molar Mass Water = 18 g/mol.
$n_{glu} = 10/180 = 0.0555 \text{ mol}$.
$n_{wat} = 90/18 = 5.0 \text{ mol}$.

2. Molality (m)

$$m = \frac{0.0555}{0.090 \text{ kg}} = 0.617 \text{ m}$$

3. Mole Fraction (x)

$$x_{glu} = \frac{0.0555}{0.0555 + 5.0} = \frac{0.0555}{5.0555} \approx 0.011$$ $$x_{wat} = 1 – 0.011 = 0.989$$

4. Molarity (M)

Vol of 100g solution = $100 / 1.2 = 83.33 \text{ mL} = 0.0833 \text{ L}$.

$$M = \frac{0.0555}{0.0833} = 0.67 \text{ M}$$

Molality = 0.617 m, $x_{glu} = 0.011$, Molarity = 0.67 M

Question 1.6

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amounts of both?

1. Moles in Mixture

Let moles of $Na_2CO_3$ and $NaHCO_3$ be $x$.
$M(Na_2CO_3) = 106$, $M(NaHCO_3) = 84$.
$106x + 84x = 1 \implies 190x = 1 \implies x = 1/190 = 0.00526 \text{ mol}$.

2. Reaction Stoichiometry

$Na_2CO_3 + 2HCl \to 2NaCl + H_2O + CO_2$ (Needs 2 mol HCl)
$NaHCO_3 + 1HCl \to NaCl + H_2O + CO_2$ (Needs 1 mol HCl)

$$n_{HCl} = 2x + x = 3x = 3 \times 0.00526 = 0.01578 \text{ mol}$$

3. Volume of HCl

$$V = \frac{n}{M} = \frac{0.01578}{0.1} = 0.1578 \text{ L} = 157.8 \text{ mL}$$

Volume Required = 157.8 mL

Question 1.7

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Calculation

Solute in Sol 1: $300 \times 0.25 = 75 \text{ g}$.
Solute in Sol 2: $400 \times 0.40 = 160 \text{ g}$.
Total Solute: $75 + 160 = 235 \text{ g}$.
Total Solution Mass: $300 + 400 = 700 \text{ g}$.

$$\text{Mass \%} = \frac{235}{700} \times 100 = 33.57\%$$

Mass % of Solute = 33.57% (Solvent = 66.43%)

Question 1.8

An antifreeze solution is prepared from 222.6 g of ethylene glycol ($C_2H_6O_2$) and 200 g of water. Calculate the molality. If density is 1.072 g/mL, calculate molarity.

1. Moles of Glycol

Molar mass $C_2H_6O_2 = 24+6+32 = 62 \text{ g/mol}$.
$n = 222.6 / 62 = 3.59 \text{ mol}$.

2. Molality

Mass solvent (water) = 200g = 0.2 kg.

$$m = \frac{3.59}{0.2} = 17.95 \text{ m}$$

3. Molarity

Total mass = $222.6 + 200 = 422.6 \text{ g}$.
Volume = $422.6 / 1.072 = 394.2 \text{ mL} = 0.3942 \text{ L}$.

$$M = \frac{3.59}{0.3942} = 9.11 \text{ M}$$

Molality = 17.95 m, Molarity = 9.11 M

Question 1.9

Drinking water contaminated with chloroform ($CHCl_3$) is 15 ppm (by mass). (i) Express in percent by mass. (ii) Determine molality.

(i) Percent by Mass

15 ppm means 15g in $10^6$ g solution.

$$\% = \frac{15}{10^6} \times 100 = 1.5 \times 10^{-3} \%$$

(ii) Molality

Since amount is very small, mass of solvent $\approx$ mass of solution = $10^6 \text{ g} = 10^3 \text{ kg}$.
Molar mass $CHCl_3 = 12+1+3(35.5) = 119.5 \text{ g/mol}$.
Moles $n = 15 / 119.5 = 0.1255 \text{ mol}$.

$$m = \frac{0.1255}{1000 \text{ kg}} = 1.255 \times 10^{-4} \text{ m}$$

Question 1.10

What role does the molecular interaction play in a solution of alcohol and water?

Alcohol and water both possess strong intermolecular Hydrogen Bonding. When mixed:

The interaction between alcohol and water molecules (A-B) is slightly weaker than the interaction between alcohol-alcohol (A-A) and water-water (B-B) molecules.
This leads to a Positive Deviation from Raoult’s Law.
The solution volume increases slightly ($\Delta V_{mix} > 0$) and mixing is endothermic ($\Delta H_{mix} > 0$).

Question 1.11

Why do gases always tend to be less soluble in liquids as the temperature is raised?

The dissolution of a gas in a liquid is an exothermic process ($\Delta H_{dissolution} < 0$) because gas molecules lose kinetic energy and form bonds with the solvent.

According to Le Chatelier’s Principle, increasing the temperature favors the endothermic direction (reverse process). Thus, the gas escapes from the liquid, and solubility decreases.

Question 1.12

State Henry’s law and mention some important applications.

Henry’s Law: The partial pressure of the gas in vapor phase ($p$) is proportional to the mole fraction of the gas ($x$) in the solution. $p = K_H x$.

Applications:

Soda Bottles: Sealed under high pressure to increase solubility of $CO_2$.
Scuba Diving: Tanks diluted with Helium to prevent “The Bends” (release of nitrogen bubbles in blood upon surfacing).
High Altitude: Low partial pressure of oxygen leads to low concentration in blood, causing Anoxia (weakness/confusion).

Question 1.13

The partial pressure of ethane over a solution containing $6.56 \times 10^{-3}$ g of ethane is 1 bar. If the solution contains $5.00 \times 10^{-2}$ g of ethane, what shall be the partial pressure?

Using Henry’s Law Relation

Since $p \propto x$ and $x \propto \text{mass}$ (for dilute solutions), we can say $p \propto m$.

$$\frac{p_1}{m_1} = \frac{p_2}{m_2}$$ $$\frac{1 \text{ bar}}{6.56 \times 10^{-3}} = \frac{p_2}{5.00 \times 10^{-2}}$$ $$p_2 = \frac{5.00 \times 10^{-2}}{6.56 \times 10^{-3}} = \frac{50}{6.56} \approx 7.62 \text{ bar}$$

Partial Pressure = 7.62 bar

Question 1.14

What is meant by positive and negative deviations from Raoult’s law and how is the sign of $\Delta_{mix}H$ related to them?

Positive Deviation: Interaction A-B is weaker than A-A and B-B. Vapor pressure is higher than predicted.
$\Delta_{mix}H > 0$ (Endothermic). Example: Ethanol + Acetone.
Negative Deviation: Interaction A-B is stronger than A-A and B-B. Vapor pressure is lower than predicted.
$\Delta_{mix}H < 0$ (Exothermic). Example: Phenol + Aniline.

Question 1.15

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Given

At normal boiling point, $P^0 = 1.013 \text{ bar}$ (1 atm).
$P_s = 1.004 \text{ bar}$.
Mass solute $w_2 = 2 \text{ g}$, Mass water $w_1 = 98 \text{ g}$. $M_1 = 18$.

Formula

Relative lowering of Vapor Pressure: $\frac{P^0 – P_s}{P^0} = \frac{n_2}{n_1} = \frac{w_2 M_1}{w_1 M_2}$.

$$\frac{1.013 – 1.004}{1.013} = \frac{2 \times 18}{98 \times M_2}$$ $$\frac{0.009}{1.013} = \frac{36}{98 M_2}$$ $$M_2 = \frac{36 \times 1.013}{98 \times 0.009} \approx 41.35 \text{ g/mol}$$

Molar Mass $\approx 41.35 \text{ g/mol}$

Question 1.16

Heptane and octane form an ideal solution. At 373 K, $P^0_{hept} = 105.2 \text{ kPa}$ and $P^0_{oct} = 46.8 \text{ kPa}$. Calculate VP of mixture of 26.0 g heptane and 35 g octane.

Calculation

$M_{hept} (C_7H_{16}) = 100$, $M_{oct} (C_8H_{18}) = 114$.
$n_H = 26/100 = 0.26$. $n_O = 35/114 = 0.307$.
Total moles = 0.567.
$x_H = 0.26/0.567 = 0.458$, $x_O = 0.542$.

Total Pressure $P = x_H P^0_H + x_O P^0_O$.

$$P = (0.458 \times 105.2) + (0.542 \times 46.8)$$ $$P = 48.18 + 25.36 = 73.54 \text{ kPa}$$

Vapor Pressure = 73.54 kPa

Question 1.17

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Analysis

1 molal means 1 mole solute in 1000g water.
Moles of water $n_1 = 1000/18 = 55.55 \text{ mol}$. Moles solute $n_2 = 1$.
Mole fraction $x_2 = 1 / (1 + 55.55) = 1/56.55 = 0.0177$.

Relative Lowering: $\frac{P^0 – P}{P^0} = x_2$.

$$\frac{12.3 – P}{12.3} = 0.0177$$ $$12.3 – P = 0.217 \implies P = 12.08 \text{ kPa}$$

Vapor Pressure = 12.08 kPa

Question 1.18

Calculate the mass of a non-volatile solute (molar mass 40 g/mol) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Analysis

Reduce TO 80% means $P_s = 0.8 P^0$.
Lowering $\frac{P^0 – 0.8P^0}{P^0} = 0.2 = x_2$.

Moles of Octane $n_1 = 114 / 114 = 1 \text{ mol}$.
$x_2 = \frac{n_2}{n_2 + 1} = 0.2 \implies n_2 = 0.2n_2 + 0.2 \implies 0.8n_2 = 0.2 \implies n_2 = 0.25 \text{ mol}$.

$$\text{Mass} = n_2 \times M_2 = 0.25 \times 40 = 10 \text{ g}$$

Mass Required = 10 g

Question 1.19

A solution containing 30 g of non-volatile solute in 90 g water has VP 2.8 kPa. Further 18 g water added, VP becomes 2.9 kPa. Calculate (i) Molar mass of solute (ii) VP of water.

Calculation

Let Molar Mass be $M$. $P^0$ be VP of pure water.
Case 1: 30g solute, 90g water ($5$ mol).
$\frac{P^0 – 2.8}{P^0} = \frac{30/M}{5} = \frac{6}{M}$ (Approx for dilute, or use exact $\frac{n_2}{n_1+n_2}$).
Using $\frac{P^0 – P}{P} = \frac{n_2}{n_1}$ (Alternative form): $\frac{P^0 – 2.8}{2.8} = \frac{30/M}{5} = \frac{6}{M}$.

Case 2: 30g solute, 108g water ($6$ mol).
$\frac{P^0 – 2.9}{2.9} = \frac{30/M}{6} = \frac{5}{M}$.

Dividing eq 1 by eq 2: $\frac{P^0 – 2.8}{2.8} \times \frac{2.9}{P^0 – 2.9} = \frac{6}{5}$.
Solving yields $P^0 \approx 3.4 \text{ kPa}$.
Sub back to find $M \approx 23 \text{ g/mol}$.

Molar Mass = 23 g/mol, $P^0$ = 3.4 kPa

Question 1.20

A 5% solution of cane sugar ($M=342$) has freezing point 271K. Calculate FP of 5% glucose ($M=180$) in water. (Pure water FP = 273.15 K).

1. Sugar Solution

$\Delta T_f = 273.15 – 271 = 2.15 \text{ K}$.
$\Delta T_f = K_f \times m$.
$m = \frac{5/342}{0.095}$. So $2.15 = K_f \frac{5}{342 \times 0.095}$.

2. Glucose Solution

$\Delta T’_f = K_f \frac{5}{180 \times 0.095}$.

Ratio $\frac{\Delta T’_f}{2.15} = \frac{342}{180} = 1.9$.
$\Delta T’_f = 2.15 \times 1.9 = 4.085 \text{ K}$.

Freezing Point = $273.15 – 4.085 = 269.065 \text{ K}$.

Freezing Point = 269.07 K

Question 1.21

Calculate atomic masses of A and B given data for $AB_2$ and $AB_4$ in benzene.

Calculation

Formula: $M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$.

For $AB_2$: $M_{AB2} = \frac{5.1 \times 1 \times 1000}{2.3 \times 20} = 110.87 \text{ g/mol}$.

For $AB_4$: $M_{AB4} = \frac{5.1 \times 1 \times 1000}{1.3 \times 20} = 196.15 \text{ g/mol}$.

$A + 2B = 110.87$ and $A + 4B = 196.15$.
Subtracting: $2B = 85.28 \implies B = 42.64$.
$A = 110.87 – 85.28 = 25.59$.

A = 25.59 u, B = 42.64 u

Question 1.22

At 300 K, 36 g of glucose per litre has osmotic pressure 4.98 bar. If osmotic pressure is 1.52 bar, what is the concentration?

Logic

$\pi = CRT$. Since T is constant, $\pi \propto C$.

$$\frac{C_2}{C_1} = \frac{\pi_2}{\pi_1}$$ $$C_1 = \frac{36}{180} = 0.2 \text{ M}$$ $$C_2 = 0.2 \times \frac{1.52}{4.98} \approx 0.061 \text{ M}$$

Concentration = 0.061 M

Question 1.23

Suggest the most important type of intermolecular attractive interaction in: (i) n-hexane and n-octane, (ii) I2 and CCl4, (iii) NaClO4 and water, (iv) methanol and acetone, (v) acetonitrile and acetone.

n-hexane and n-octane: London Dispersion Forces (Both non-polar).
I2 and CCl4: London Dispersion Forces (Both non-polar).
NaClO4 and water: Ion-Dipole Interaction (Ionic solid in polar solvent).
Methanol and acetone: Hydrogen Bonding (Dipole-Dipole).
Acetonitrile and acetone: Dipole-Dipole Interaction.

Question 1.24

Based on solute-solvent interactions, arrange in increasing solubility in n-octane: Cyclohexane, KCl, CH3OH, CH3CN.

n-octane is non-polar. “Like dissolves like”. Non-polar solutes will be most soluble.

Order: KCl < CH3OH < CH3CN < Cyclohexane.

Reason: KCl (Ionic) < CH3OH (Polar/H-bond) < CH3CN (Polar) < Cyclohexane (Non-polar).

Question 1.25

Identify insoluble, partially soluble, and highly soluble in water for: phenol, toluene, formic acid, ethylene glycol, chloroform, pentanol.

Insoluble: Toluene, Chloroform (Non-polar/Weak polar).
Partially Soluble: Phenol, Pentanol (Polar -OH group but large non-polar chain/ring).
Highly Soluble: Formic acid, Ethylene glycol (Strong H-bonding).

Question 1.26

Lake water ($d=1.25 \text{ g/mL}$) contains 92 g $Na^+$ per kg of water. Calculate molarity of $Na^+$.

Calculation

Moles $Na^+ = 92 / 23 = 4 \text{ mol}$.
Mass of solution = $1000 \text{ g water} + 92 \text{ g ions} = 1092 \text{ g}$.
Volume = $1092 / 1.25 = 873.6 \text{ mL} = 0.8736 \text{ L}$.

$$M = \frac{4}{0.8736} \approx 4.58 \text{ M}$$

Molarity = 4.58 M

Question 1.27

If solubility product of CuS is $6 \times 10^{-16}$, calculate maximum molarity.

Calculation

$CuS \rightleftharpoons Cu^{2+} + S^{2-}$.
$K_{sp} = S \times S = S^2$.

$$S = \sqrt{6 \times 10^{-16}} = 2.45 \times 10^{-8} \text{ M}$$

Molarity = $2.45 \times 10^{-8} \text{ M}$

Question 1.28

Calculate mass % of aspirin ($C_9H_8O_4$) in acetonitrile when 6.5 g is dissolved in 450 g.

$$\text{Mass \%} = \frac{6.5}{6.5 + 450} \times 100 = \frac{6.5}{456.5} \times 100 \approx 1.42\%$$

Mass Percentage = 1.42%

Question 1.29

Nalorphene ($C_{19}H_{21}NO_3$), dose 1.5 mg. Calculate mass of $1.5 \times 10^{-3} \text{ m}$ aqueous solution required.

Calculation

Molar Mass = 311 g/mol.
$1.5 \times 10^{-3} \text{ m}$ means $1.5 \times 10^{-3}$ moles in 1000g water.
Mass of solute = $1.5 \times 10^{-3} \times 311 = 0.4665 \text{ g}$.
Mass of solution = $1000 + 0.4665 = 1000.4665 \text{ g}$.

We need 1.5 mg ($1.5 \times 10^{-3} \text{ g}$) of solute.
Mass solution required = $\frac{1000.4665}{0.4665} \times 1.5 \times 10^{-3} \approx 3.22 \text{ g}$.

Mass of Solution = 3.22 g

Question 1.30

Calculate amount of benzoic acid ($C_6H_5COOH$) for 250 mL of 0.15 M solution.

Calculation

Molar Mass = 122 g/mol.
Moles = $M \times V = 0.15 \times 0.25 = 0.0375 \text{ mol}$.
Mass = $0.0375 \times 122 = 4.575 \text{ g}$.

Mass = 4.575 g

Question 1.31

Depression in freezing point increases: Acetic acid < Trichloroacetic < Trifluoroacetic. Explain.

Fluorine is more electronegative than Chlorine. Electron withdrawing groups increase the acidity (ionization) of the carboxylic acid.
Stronger acid $\to$ Higher dissociation ($\alpha$) $\to$ Higher van’t Hoff factor ($i$) $\to$ Higher $\Delta T_f$.

Question 1.32

Calculate depression in freezing point: 10g $CH_3CH_2CHClCOOH$ in 250g water. $K_a = 1.4 \times 10^{-3}$.

1. Molality

Molar Mass = 122.5 g/mol. $n = 10/122.5 = 0.0816$.
$m = 0.0816 / 0.25 = 0.326 \text{ m}$.

2. Dissociation

$\alpha = \sqrt{K_a/m} = \sqrt{1.4 \times 10^{-3} / 0.326} \approx 0.0655$.
$i = 1 + \alpha = 1.0655$.

3. Depression

$\Delta T_f = i K_f m = 1.0655 \times 1.86 \times 0.326 \approx 0.65 \text{ K}$.

Depression = 0.65 K

Question 1.33

19.5g $CH_2FCOOH$ in 500g water. $\Delta T_f = 1.0^\circ C$. Calculate $i$ and $K_a$.

1. van’t Hoff Factor

$M = 78 \text{ g/mol}$. $m = (19.5/78)/0.5 = 0.5 \text{ m}$.
Calculated $\Delta T_f = 1.86 \times 0.5 = 0.93$.
$i = \text{Observed} / \text{Calculated} = 1.0 / 0.93 = 1.0753$.

2. Dissociation Constant

$\alpha = i – 1 = 0.0753$.
$K_a = \frac{C \alpha^2}{1-\alpha} \approx m \alpha^2$ (approx).
$K_a = 0.5 \times (0.0753)^2 \approx 2.83 \times 10^{-3}$ (exact calc yields $3.07 \times 10^{-3}$).

$i = 1.075$, $K_a \approx 3.07 \times 10^{-3}$.

Question 1.34

VP of water 17.535 mm Hg. 25g glucose in 450g water. Calculate VP.

Calculation

$n_2 = 25/180 = 0.139$. $n_1 = 450/18 = 25$.
$x_2 = 0.139 / 25.139 = 0.0055$.
Lowering = $P^0 x_2 = 17.535 \times 0.0055 = 0.096$.
$P = 17.535 – 0.096 = 17.44 \text{ mm Hg}$.

Vapor Pressure = 17.44 mm Hg

Question 1.35

Henry’s law constant for methane in benzene is $4.27 \times 10^5 \text{ mm Hg}$. Calculate solubility at 760 mm Hg.

Calculation

$P = K_H x \implies x = P / K_H$.
$x = 760 / (4.27 \times 10^5) = 178 \times 10^{-5} = 1.78 \times 10^{-3}$.

Solubility (mole fraction) = $1.78 \times 10^{-3}$

Question 1.36

100g A ($M=140$) in 1000g B ($M=180$). $P_B^0 = 500$. $P_{total} = 475$. Calc $P_A^0$ and $P_A$.

1. Moles

$n_A = 100/140 = 0.714$. $n_B = 1000/180 = 5.556$.
$x_A = 0.114$. $x_B = 0.886$.

2. Pressure

$P_{total} = P_A + P_B = P_A^0 x_A + P_B^0 x_B$.
$475 = P_A^0(0.114) + 500(0.886)$.
$475 = 0.114 P_A^0 + 443$.
$32 = 0.114 P_A^0 \implies P_A^0 = 280.7 \text{ Torr}$.
$P_A = 280.7 \times 0.114 = 32 \text{ Torr}$.

$P_A^0 = 280.7 \text{ Torr}$, $P_A = 32 \text{ Torr}$

Question 1.37

Plot data for Acetone and Chloroform. Indicate deviation.

Analysis: Acetone and Chloroform form hydrogen bonds with each other, which are stronger than the individual intermolecular forces. This results in a decrease in vapor pressure.

Result: Negative Deviation from Raoult’s Law.

Question 1.38

Benzene and Toluene. $P_B^0 = 50.71$, $P_T^0 = 32.06$. 80g Benzene, 100g Toluene. Calc mole fraction of benzene in vapor phase.

1. Liquid Composition

$n_B = 80/78 = 1.026$. $n_T = 100/92 = 1.087$.
$x_B = 0.486$. $x_T = 0.514$.

2. Partial Pressures

$P_B = 50.71 \times 0.486 = 24.65$.
$P_T = 32.06 \times 0.514 = 16.48$.
$P_{total} = 41.13$.

3. Vapor Phase

$y_B = P_B / P_{total} = 24.65 / 41.13 = 0.60$.

Mole fraction in vapor = 0.60

Question 1.39

Air (20% O2, 79% N2). P=10 atm. Calc composition in water. $K_H(O_2) = 3.3 \times 10^7 \text{ mm}$, $K_H(N_2) = 6.51 \times 10^7 \text{ mm}$.

Calculation

Total P = 10 atm = 7600 mm Hg.
$p_{O2} = 0.2 \times 7600 = 1520 \text{ mm}$.
$p_{N2} = 0.79 \times 7600 = 6004 \text{ mm}$.

$x_{O2} = 1520 / (3.3 \times 10^7) = 4.6 \times 10^{-5}$.
$x_{N2} = 6004 / (6.51 \times 10^7) = 9.2 \times 10^{-5}$.

$x_{O2} = 4.6 \times 10^{-5}$, $x_{N2} = 9.2 \times 10^{-5}$

Question 1.40

Amount of $CaCl_2$ ($i=2.47$) in 2.5L water for $\pi = 0.75 \text{ atm}$ at 27 C.

Calculation

$\pi = iCRT = i \frac{n}{V} RT$.
$n = \frac{\pi V}{i R T} = \frac{0.75 \times 2.5}{2.47 \times 0.0821 \times 300}$.
$n = 1.875 / 60.83 \approx 0.0308 \text{ mol}$.
Mass = $0.0308 \times 111 = 3.42 \text{ g}$.

Mass = 3.42 g

Question 1.41

Osmotic pressure of 25 mg $K_2SO_4$ in 2L water. Complete dissociation.

Calculation

$K_2SO_4 \to 2K^+ + SO_4^{2-}$, so $i = 3$.
Moles $n = 0.025 / 174 \approx 1.44 \times 10^{-4}$.
$\pi = 3 \times \frac{1.44 \times 10^{-4}}{2} \times 0.0821 \times 298 \approx 5.27 \times 10^{-3} \text{ atm}$.

Osmotic Pressure = $5.27 \times 10^{-3} \text{ atm}$

NCERT Solutions