CLASS 12 · CHEMISTRY · CHAPTER 1

Solutions

Simple and explained answers for all 41 end-of-chapter questions on solution concentration, gas solubility, vapour pressure, and colligative properties.

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Chapter 1 Exercises

Textbook pages 27–30

Question 1

Define a solution and describe the main types with examples.

Steps only

A solution is a uniform mixture; its state may be gas, liquid, or solid.

Answer

A solution is a homogeneous mixture. It may be gaseous (air), liquid (sugar dissolved in water), or solid (copper in gold).

Question 2

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

Steps only

Hydrogen in palladium.

Answer

Hydrogen dissolved in palladium is a solid solution with a gas as the solute.

Question 3

Explain mole fraction, molality, molarity, and mass percentage.

Steps only

Use the amount of substance for the numerator; molality uses kg of solvent, while molarity uses L of solution.

Answer

Mole fraction: component moles / total moles. Molality: solute moles per kg solvent. Molarity: solute moles per litre solution. Mass percentage: component mass / solution mass × 100.

Question 4

Find the molarity of 68% by-mass HNO₃ of density 1.504 g mL⁻¹.

Steps only

In 1 L: solution mass = 1504 g; HNO₃ = 1022.72 g = 16.23 mol. So M = 16.23 M.

Answer

Molarity ≈ 16.2 mol L⁻¹ (16.23 M).

Question 5

For 10% w/w glucose solution, find molality, both mole fractions, and molarity if density is 1.2 g mL⁻¹.

Steps only

Use a 100 g basis, convert both components to moles, then use the solution density to find its volume.

Answer

Molality = 0.617 m; mole fraction glucose = 0.0110 and water = 0.9890; molarity = 0.667 M.

Question 6

A 1 g equimolar mixture of Na₂CO₃ and NaHCO₃ is titrated with 0.1 M HCl. Find the required volume.

Steps only

Equimolar salt amounts are each 0.005263 mol; HCl needed = 3 times this = 0.01579 mol = 158 mL.

Answer

About 158 mL of 0.1 M HCl.

Question 7

Mix 300 g of 25% solution with 400 g of 40% solution. Find the final mass percentage.

Steps only

Add solute masses and divide by the combined solution mass: 235/700 × 100 = 33.6%.

Answer

The resulting solution is 33.6% by mass.

Question 8

Find molality and molarity for 222.6 g ethylene glycol in 200 g water; solution density is 1.072 g mL⁻¹.

Steps only

Ethylene glycol = 3.59 mol. Divide by 0.200 kg water for molality; use density to get 0.3942 L solution for molarity.

Answer

Molality ≈ 17.95 m; molarity ≈ 9.11 M.

Question 9

A water sample contains 15 ppm by mass of CHCl₃. Convert to mass percent and find molality.

Steps only

15 ppm = 0.0015%. In 1 kg, 0.015 g CHCl₃ gives about 1.26 × 10⁻⁴ mol kg⁻¹.

Answer

Mass percentage = 0.0015%; molality ≈ 1.26 × 10⁻⁴ mol kg⁻¹.

Question 10

What happens to molecular interactions when alcohol and water are mixed?

Steps only

They form hydrogen bonds, but the mixed attractions are not identical to the pure-liquid attractions, so the mixture is non-ideal.

Answer

Alcohol and water form hydrogen bonds, but the unlike interactions differ from those in the pure liquids. The mixture is non-ideal; ethanol-water shows positive deviation from Raoult’s law.

Question 11

Why does gas solubility in a liquid usually decrease as temperature rises?

Steps only

Gas dissolution is usually exothermic; heating favours gas leaving the liquid.

Answer

Gas dissolution usually releases heat. Heating shifts the process towards gas escaping, so solubility decreases.

Question 12

State Henry’s law and give applications.

Steps only

At constant temperature, p = Kₕx. High pressure carbonates drinks; pressure changes explain diving-related dissolved-gas effects.

Answer

Henry’s law: p = Kₕx at constant temperature. Applications include carbonating drinks and understanding gas absorption in divers’ blood.

Question 13

Ethane pressure is 1 bar at 6.56 × 10⁻³ g dissolved ethane. Estimate pressure for 5.00 × 10⁻² g in the same solution conditions.

Steps only

Pressure scales with dissolved amount: (0.0500/0.00656) × 1 bar = 7.62 bar.

Answer

The partial pressure is approximately 7.62 bar.

Question 14

Explain positive and negative deviations from Raoult’s law and relate them to mixing enthalpy.

Steps only

Weaker unlike attractions give positive deviation and endothermic mixing; stronger unlike attractions give negative deviation and exothermic mixing.

Answer

Positive deviation: weaker A–B attractions and ΔmixH > 0. Negative deviation: stronger A–B attractions and ΔmixH < 0.

Question 15

A 2% aqueous non-volatile solute solution has pressure 1.004 bar at water’s normal boiling point. Estimate solute molar mass.

Steps only

Relative pressure lowering ≈ 0.00888; water = 5.444 mol, so solute ≈ 0.04835 mol in 2 g: M ≈ 41.4 g mol⁻¹.

Answer

Molar mass ≈ 41 g mol⁻¹.

Question 16

Calculate total vapour pressure for 26.0 g heptane and 35 g octane at 373 K. Pure vapour pressures: 105.2 and 46.8 kPa.

Steps only

Find mole fractions from the masses, then p_total = (0.458)(105.2) + (0.542)(46.8) ≈ 73.5 kPa.

Answer

Total vapour pressure ≈ 73.5 kPa.

Question 17

Water has vapour pressure 12.3 kPa at 300 K. Find it for a 1 molal non-volatile solute solution.

Steps only

x_water = 55.56/(56.56) = 0.9823; p = 0.9823 × 12.3 = 12.1 kPa.

Answer

Vapour pressure ≈ 12.1 kPa.

Question 18

Find the mass of a non-volatile 40 g mol⁻¹ solute that lowers octane’s vapour pressure to 80% of pure octane; 114 g octane is used.

Steps only

One mole octane with x_octane = 0.80 requires 0.25 mol solute, or 10 g.

Answer

Add 10 g of solute.

Question 19

A solution has 30 g solute in 90 g water, pressure 2.8 kPa. After adding 18 g water, pressure is 2.9 kPa. Find solute molar mass and pure-water pressure.

Steps only

Using Raoult’s law for both states gives n ≈ 0.714 mol and p° ≈ 3.20 kPa; M = 30/0.714 ≈ 42 g mol⁻¹.

Answer

Solute molar mass ≈ 42 g mol⁻¹; pure-water vapour pressure ≈ 3.20 kPa.

Question 20

A 5% sugar solution freezes at 271 K. Estimate the freezing point of 5% glucose solution; pure water freezes at 273.15 K.

Steps only

Depression scales with moles: 2.15 × (5/180)/(5/342) = 4.09 K; 273.15−4.09 = 269.06 K.

Answer

Freezing point ≈ 269.1 K.

Question 21

Elements A and B form AB₂ and AB₄. From their freezing-point depressions in benzene, determine atomic masses.

Steps only

The apparent molar masses are 110.87 and 196.15. Solving A+2B and A+4B gives A≈25.6, B≈42.6.

Answer

A ≈ 25.6 u and B ≈ 42.6 u (about 26 u and 43 u).

Question 22

A 1 L solution containing 36 g glucose has osmotic pressure 4.98 bar at 300 K. What concentration gives 1.52 bar at the same temperature?

Steps only

C₂/C₁ = Π₂/Π₁; 36 × 1.52/4.98 = 11.0 g L⁻¹ = 0.0611 M.

Answer

Concentration ≈ 11.0 g L⁻¹, or 0.0611 M.

Question 23

Identify the main attraction for each pair: (i) n-hexane/n-octane; (ii) I₂/CCl₄; (iii) NaClO₄/water; (iv) methanol/acetone; (v) acetonitrile/acetone.

Steps only

Non-polar pairs: dispersion; salt-water: ion-dipole; methanol-acetone: hydrogen bonding; two polar molecules: dipole-dipole.

Answer

(i) London dispersion;
(ii) London dispersion;
(iii) ion-dipole;
(iv) hydrogen bonding;
(v) dipole-dipole.

Question 24

Arrange cyclohexane, KCl, methanol, and acetonitrile by increasing solubility in n-octane.

Steps only

“Like dissolves like”: ionic and polar compounds dissolve poorly in non-polar octane; non-polar cyclohexane dissolves best.

Answer

KCl < CH₃OH < CH₃CN < cyclohexane.

Question 25

Classify phenol, toluene, formic acid, ethylene glycol, chloroform, and pentanol by water solubility.

Steps only

Strongly hydrogen-bonding compounds are highly soluble; phenol and pentanol are partial; toluene and chloroform are insoluble/very sparingly soluble.

Answer

Highly soluble: formic acid, ethylene glycol. Partially soluble: phenol, pentanol. Insoluble: toluene, chloroform.

Question 26

Lake water density is 1.25 g mL⁻¹ and contains 92 g Na⁺ per kg of water. Find Na⁺ molarity.

Steps only

4 mol Na⁺ are in about 0.8736 L solution; M ≈ 4.58 M.

Answer

Na⁺ molarity ≈ 4.58 mol L⁻¹.

Question 27

CuS has Ksp = 6 × 10⁻¹⁶. Estimate its maximum molarity in pure water.

Steps only

For 1:1 CuS, Ksp=s²; s=√(6×10⁻¹⁶)=2.45×10⁻⁸ M.

Answer

Maximum molarity ≈ 2.45 × 10⁻⁸ M.

Question 28

Find mass percentage of 6.5 g aspirin dissolved in 450 g acetonitrile.

Steps only

6.5/(456.5) × 100 = 1.42%.

Answer

Aspirin = 1.42% by mass.

Question 29

A 1.5 × 10⁻³ m aqueous solution supplies a 1.5 mg dose of nalorphene (molar mass 311 g mol⁻¹). Find solution mass.

Steps only

Moles in 1.5 mg = 4.82×10⁻⁶; divide by 1.5×10⁻³ mol kg⁻¹ to get ≈3.21 g water/solution.

Answer

Approximately 3.21 g of solution is required.

Question 30

Find benzoic acid mass for 250 mL of 0.15 M solution in methanol.

Steps only

n=0.15×0.250=0.0375 mol; mass=0.0375×122≈4.58 g.

Answer

Use about 4.58 g benzoic acid.

Question 31

Why does freezing-point depression increase from acetic acid to trichloroacetic acid to trifluoroacetic acid?

Steps only

Acid strength and ionisation rise in the order acetic < trichloroacetic < trifluoroacetic, so particle count and ΔTf rise.

Answer

Electron-withdrawing halogens strengthen the acid; fluorine does this more strongly than chlorine. Greater ionisation creates more particles and increases freezing-point depression.

Question 32

Find freezing-point depression for 10 g CH₃CH₂CHClCOOH in 250 g water; Ka=1.4×10⁻³ and Kf=1.86 K kg mol⁻¹.

Steps only

m≈0.3265; weak-acid dissociation α≈0.063, so i≈1.063. ΔTf=iKfm≈0.646 K.

Answer

Freezing-point depression ≈ 0.65 K.

Question 33

For 19.5 g fluoroacetic acid (CH₂FCOOH) in 500 g water, observed ΔTf=1.00°C. Find i and Ka.

Steps only

m=0.50; i=1/0.93=1.075; α≈0.075 and Ka≈mα²/(1−α)=3.0×10⁻³.

Answer

i ≈ 1.075; Ka ≈ 3.0 × 10⁻³.

Question 34

Water vapour pressure at 293 K is 17.535 mm Hg. Find it after dissolving 25 g glucose in 450 g water.

Steps only

x_water=25/(25+0.1389)=0.99448; p=0.99448×17.535≈17.44 mm Hg.

Answer

Vapour pressure ≈ 17.44 mm Hg.

Question 35

Methane has Henry constant 4.27 × 10⁵ mm Hg (molality form). Find its molality under 760 mm Hg.

Steps only

m = 760/(4.27×10⁵) = 1.78×10⁻³ mol kg⁻¹.

Answer

Methane solubility ≈ 1.78 × 10⁻³ mol kg⁻¹.

Question 36

100 g A (M=140) is mixed with 1000 g B (M=180). Pure B pressure is 500 torr; total solution pressure is 475 torr. Find pure A pressure and A partial pressure.

Steps only

xA≈0.114; pB≈443 torr; pA=475−443≈32 torr; pA°=32/0.114≈281 torr.

Answer

Pure A vapour pressure ≈ 281 torr; A partial pressure in solution ≈ 32 torr.

Question 37

Describe how to plot ideal and observed vapour-pressure curves for acetone-chloroform data, and state the deviation.

Steps only

Ideal: pA=741.8x, pC=632.8(1−x). Observed points lie below the ideal total curve: negative deviation.

Answer

Plot the ideal Raoult-law lines and the supplied observed points against x_acetone. The data show negative deviation from Raoult’s law.

Question 38

Find vapour-phase benzene mole fraction when 80 g benzene and 100 g toluene form an ideal solution at 300 K. Pure pressures: 50.71 and 32.06 mm Hg.

Steps only

Liquid x_benzene≈0.486; partial pressures ≈24.65 and 16.47 mm Hg, so y_benzene≈0.600.

Answer

Mole fraction of benzene in the vapour ≈ 0.600.

Question 39

Air at 10 atm contains 20% O₂ and 79% N₂. Find approximate dissolved-gas composition using Henry constants O₂=3.30×10⁷ mm Hg and N₂=6.51×10⁷ mm Hg.

Steps only

Use partial pressure divided by Henry constant. The resulting mole fractions are 4.61×10⁻⁵ (O₂) and 9.22×10⁻⁵ (N₂).

Answer

Dissolved mole fractions: xO₂≈4.61×10⁻⁵ and xN₂≈9.22×10⁻⁵. Among dissolved O₂+N₂, about 33.3% is O₂ and 66.7% is N₂.

Question 40

Find the CaCl₂ mass for osmotic pressure 0.75 atm in 2.5 L at 27°C; i=2.47.

Steps only

C=Π/(iRT)=0.01232 M; n=0.01232×2.5=0.0308 mol; mass≈0.0308×111=3.42 g.

Answer

About 3.42 g CaCl₂ is required.

Question 41

Find osmotic pressure for 25 mg K₂SO₄ in 2 L water at 25°C, assuming complete dissociation.

Steps only

n=0.025/174 mol; C=n/2; Π=3CRT≈0.00527 atm≈534 Pa.

Answer

Osmotic pressure ≈ 5.27 × 10⁻³ atm (about 534 Pa).