Calculate molality of 2 . 5 g of ethanoic acid CH 3 COOH in 75 g of benzene.

Molar mass of ethanoic acid = 60 g mol - 1 Mass of solvent(benzene) = 75 g = 0 . 075 Kg Molality = ? Molality ( m ) = 2 . 5 60 × 1 0 . 075 m = 2 . 5 60 × 1 0 . 075 m = 0 . 55 mol Kg - 1

1 litre of water under a nitrogen pressure of 1 atm dissolves 2 × 10 - 5 kg of nitrogen at 293 K . Calculate Henry's law constant.

For water, 1 kg = 1 L 2 × 10 - 5 kg ( or 0 . 02 g ) nitrogen corresponds to 0 .02 28 = 7 .2 × 10 − 4 moles . Hence, the solubility s = 7 .2 × 10 − 4 M According to Henry’s law, s = K H . p s is the solubility in moles per litre K H is the henry’s law constant p is the pressure in atm. Substitute values in the above expression. 7 .2 × 10 − 4 = K H × 1 atm K H = 7 .2 × 10 − 4 L atm − 1

The Henry's law constant for CO 2 in water at 298 K is 1 . 67 k bar . Calculate the solubility of CO 2 at 298 K when the pressure of CO 2 is one bar.

We know Henry’s Law: P = K H × x where P is pressure, K H is Henry's constant and x is mole fraction of solute. ⇒ x = 1 × 10 5 1 .67 × 10 8 1 bar = 1 × 10 5 N m 2 ⇒ n CO 2 1000 / 18 = 1 × 10 5 1 .67 × 10 8 As x = n CO 2 1000 / 18 n CO 2 < < < 1 ⇒ n CO 2 = 1 × 10 5 × 10 3 1 .67 × 10 8 × 18 ⇒ n CO 2 = 0 . 033 Solubility ( or Molarity) = n CO 2 V solution Solubility = 0 . 033 1 ⇒ 3 . 3 × 10 - 2 mol L - 1

Vapour pressure of chloroform CHCl 3 and dichloromethane CH 2 Cl 2 at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate mole fractions of each component in vapour phase.

Vapour pressure of pure chloroform = P A ∘ = 200 mmHg Vapour pressure of pure dichloromethane P B o = 415 mmHg Molar mass M A = 119 . 5 g Molar mass M B = 85 g mass of CHCl 3 = 25 . 5 g Number of mole = 25 .5 119 .5 = 0 .21 Mass of CH 2 Cl 2 = 40 g Number of mole = 40 85 = 0.47 x A = 0 .213 0 .213 + 0 .47 = 0 .322 x B = 1 − 0 .322 = 0 .688 P A = P A o X A = 200 × 0 . 312 = 62 . 4 P B = P B o X B = 0 . 688 × 415 = 285 . 52 P = 62 . 4 + 285 . 52 = 347 . 92 Mole fraction of CH 2 Cl 2 in vapour phase = 285 . 52 347 . 92 ≃ 0 . 82 Mole fraction of CHCl 3 in vapour phase = 62 . 4 347 . 92 ≃ 0 . 18

Write two differences between ideal and non-ideal solutions.

Ideal solutions : An ideal solution is the solution which obeys Raoult's law perfectly over entire range of concentration and temperature and during its formation there is no change in its enthalpy and volume. Such solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecualr forces. Examples of ideal solution:- Chlorobenzene and bromobenzene; Benzene and toluene; Ethyl bromide and ethyl iodide. Non-ideal solutions : Those solutions which do not obey Rault's law are called non-ideal solutions. Non ideal solutions are formed when the components differ much in their structures and polarities. Non ideal solutions show either positive or negative deviations from Rault's law. Examples of non-ideal solution :- Ethanol and cyclohexane; Acetone and benzene; Water and methanol.

MEDIUM

12th West Bengal Board

IMPORTANT

Earn 100

Calculate the molarity of a solution containing $5 g$ of $NaOH$ in $450 mL$ solution.

Important Questions on Solutions

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 2 - Solutions>Follow Up Problems>Q 12

Calculate molality of

2.5 g

of ethanoic acid

({CH}_{3} COOH)

75 g

of benzene.

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 2 - Solutions>Follow Up Problems>Q 13

N_{2}

gas is bubbled through water at

293 K,

how many millimoles of

N_{2}

gas would dissolve in

1

litre of water. Assume that

N_{2}

exerts a partial pressure of

0.987 bar .

Given that Henry's law constant for

N_{2}

293 K

76.48 k bar .

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 2 - Solutions>Follow Up Problems>Q 14

1

litre of water under a nitrogen pressure of

1 atm

dissolves

2 \times 10^{- 5} kg

of nitrogen at

293 K .

Calculate Henry's law constant.

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 2 - Solutions>Follow Up Problems>Q 15

The Henry's law constant for

{CO}_{2}

in water at

298 K

1.67 k bar .

Calculate the solubility of

{CO}_{2}

298 K

when the pressure of

{CO}_{2}

is one bar.

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 2 - Solutions>Follow Up Problems>Q 16

Vapour pressure of chloroform

({CHCl}_{3})

and dichloromethane

({CH}_{2} {Cl}_{2})

298 K

are

200 mm Hg

and

415 mm Hg

respectively.
Calculate the vapour pressure of the solution prepared by mixing

25.5 g

{CHCl}_{3}

and

40 g

{CH}_{2} {Cl}_{2}

298 K

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 2 - Solutions>Follow Up Problems>Q 16

Vapour pressure of chloroform

({CHCl}_{3})

and dichloromethane

({CH}_{2} {Cl}_{2})

298 K

are

200 mm Hg

and

415 mm Hg

respectively. Calculate mole fractions of each component in vapour phase.

MEDIUM

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 2 - Solutions>Follow Up Problems>Q 17

Heptane and octane form an ideal solution at

373 K .

The vapour pressure of the pure liquids at this temperature is

105.2 k Pa

and

46.8 k Pa,

respectively. If the solution contains

25 g

of heptane and

28.5 g

of octane. Calculate vapour pressure exerted by heptane.

MEDIUM

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 2 - Solutions>Follow Up Problems>Q 17

Heptane and octane form an ideal solution at

373 K .

The vapour pressure of the pure liquids at this temperature is

105.2 k Pa

and

46.8 k Pa,

respectively. If the solution contains

25 g

of heptane and

28.5 g

of octane. Calculate vapour pressure exerted by the solution.

Calculate the molarity of a solution containing 5g of NaOH in 450 mL solution.

Important Questions on Solutions

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Calculate the molarity of a solution containing $5 g$ of $NaOH$ in $450 mL$ solution.