Intext Questions

 1.1 Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Mass of benzene (C₆H₆) = 22 g
Mass of carbon tetrachloride (CCl₄) = 122 g

Step 1: Calculate total mass of solution

Total mass=22+122=144 g

1.2 Calculate the mole fraction of benzene in solution containing 30%
by mass in carbon tetrachloride

Solution contains 30% by mass benzene in carbon tetrachloride (CCl₄)

That means in 100 g of solution:

• Mass of benzene (C₆H₆) = 30 g
• Mass of carbon tetrachloride (CCl₄) = 70 g

Step 1: Calculate molar masses

• Molar mass of C₆H₆ = $6×12 + 6×1 = 78$g/mol
• Molar mass of CCl₄ = $12 + 4×35.5 = 154$g/mol

1.3 Calculate the molarity of each of the following solutions: (a) 30 g of
Co(NO₃)₂. 6H₂O in 4.3 L of solution (b) 30 mL of 0.5 M H₂SO₄ diluted to
500 mL

Calculate molar mass

For Co(NO₃)₂·6H₂O

• Co = 59
• (NO₃)₂ = 2 × (14 + 3×16) = 2 × 62 = 124
• 6H₂O = 6 × 18 = 108

Molar mass=59+124+108=291 g/mol

Step 2: Calculate moles

Moles=30/291=0.103 mol

b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL

Use dilution formula:$$M_1V_1 = M_2V_2$$​

Given:

• $M_1 = 0.5$M
• $V_1 = 30$ mL
• $V_2 = 500$ mL

1.4 Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of
0.25 molal aqueous solution

Mass of solution = 2.5 kg
Molality (m) = 0.25 mol/kg
Solute = Urea (NH₂CONH₂)

1.5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density
of 20% (mass/mass) aqueous KI is 1.202 g mL^-1

20% (w/w) aqueous KI
Density = 1.202 g mL⁻¹

That means:

In 100 g of solution

• Mass of KI = 20 g
• Mass of water = 80 g

Molar mass of KI = $39 + 127 = 166 \text{ g/mol}$

1.6 H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.

To calculate Henry’s law constant (Kₕ), we use:

p=K_H​×x

Where:

• $p$ = partial pressure of gas (at STP = 1 atm)
• $x$= mole fraction of gas in solution
• Given solubility = 0.195 molal (m)

1.7 Henry’s law constant for CO₂ in water is 1.67×10⁸ Pa at 298 K. Calculate
the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm
CO₂ pressure at 298 K.

Given:

• Henry’s constant, $K_H = 1.67 \times 10^8$
• Pressure of CO₂ = 2.5 atm
• Volume of soda water = 500 mL = 0.5 L
• Temperature = 298 K

1.8 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg
respectively, at 350 K . Find out the composition of the liquid mixture if total
vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

1.9 Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

1.10 Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

Given:

Boiling point at 750 mm Hg = 99.63°C
Required boiling point = 100°C

ΔTb​=100−99.63=0.37^∘C

Mass of water = 500 g = 0.5 kg

For water: K_b​=0.52Kkgmol⁻¹

Sucrose is non-electrolyte → $i = 1$

Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342 g/mol

1.11 Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg mol^-1

Given:

Depression in freezing point: ΔT_f​=1.5^∘C

K_f​=3.9Kkgmol⁻¹

Mass of solvent (acetic acid) = 75 g = 0.075 kg

Ascorbic acid (C₆H₈O₆) is non-electrolyte → $i = 1$

Molar mass of ascorbic acid:

6(12)+8(1)+6(16)

=72+8+96=176g/mol

1.12 Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C

Given:

Mass of polymer = 1.0 g
Molar mass = 185,000 g/mol

Volume of solution = 450 mL = 0.450 L
Temperature = 37°C = 310 K

Gas constant:

R=0.0821Latmmol⁻¹K⁻¹

Step 1: Use osmotic pressure formula

π=CRT

First find molarity (C).

Exercises

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

Definition of Solution

A solution is a homogeneous mixture of two or more components in which:

• The component present in larger amount is called solvent
• The component present in smaller amount is called solute

Example: Salt dissolved in water.

Types of Solutions

Solutions are classified on the basis of physical state of solute and solvent.

There are 9 types of solutions:

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1. Gas in Gas Solution

• Solvent: Gas
• Solute: Gas
• Example: Air (O₂, CO₂ dissolved in N₂)

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2. Gas in Liquid Solution

• Solvent: Liquid
• Solute: Gas
• Example: CO₂ in water (Soda water)

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3. Gas in Solid Solution

• Solvent: Solid
• Solute: Gas
• Example: Hydrogen in palladium

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4. Liquid in Gas Solution

• Solvent: Gas
• Solute: Liquid
• Example: Moist air (water vapour in air)

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5. Liquid in Liquid Solution

• Solvent: Liquid
• Solute: Liquid
• Example: Alcohol in water

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6. Liquid in Solid Solution

• Solvent: Solid
• Solute: Liquid
• Example: Dental amalgam (Mercury in silver)

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7. Solid in Gas Solution

• Solvent: Gas
• Solute: Solid
• Example: Smoke (dust particles in air)

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8. Solid in Liquid Solution

• Solvent: Liquid
• Solute: Solid
• Example: Salt in water

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9. Solid in Solid Solution

• Solvent: Solid
• Solute: Solid
• Example: Alloys (Brass = Zn in Cu)

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

Hydrogen gas dissolved in palladium metal is an example of a solid solution in which the solute is a gas.

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