Vapour pressure of water at 293 K is 17 . 535 mm Hg . Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Given:- P 0 = 17 . 535 mm Molar mass of glucose ( C 6 H 12 O 6 ) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1 Molar mass of water = 18 g mol - 1 Raoult ' s law p 0 - p s p 0 = n 2 n 1 + n 2 = n 2 n 1 = w 2 / M 2 w 1 / M 1 1 - p s p 0 = 25 / 180 450 / 18 = 25 × 18 180 × 450 1 - 450 81000 = p s p 0 p s = 0 . 9945 × 17 . 535 = 17 . 44 mm Hg

Raoult's law: According to this law, the partial pressure of any volatile constituents of a solution at a constant temperature is equal to the vapour pressure of pure constituents multiplied by the mole fraction of that constituent in the solution. For liquid-liquid solution: Let a mixture (solution) be prepared by mixing n A moles of liquid A and n B moles of li quid B . Let P A and P B be the partial pressures of two constituents A and B in solution and P A 0 and P B 0 the vapour pressures in pure state respectively. Thus, according Raoult's law, P A = X A P A 0 and P B = X B P B 0 .

What is an ideal solution

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 2 - Solutions>Questions>Q 40

What is relative lowering of vapour pressure? How is it useful to determine the molar mass of a solute?

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 2 - Solutions>Questions>Q 41

How is molar mass related to the depression in freezing point of a solution?

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12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 2 - Solutions>Questions>Q 42

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?

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12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 2 - Solutions>Questions>Q 43

Heptane and octane form an ideal solution. At

373 K

, the vapour pressures of the two liquid components are

105.2 kPa

and

46.8 kPa

respectively. What will be the vapour pressure of a mixture of

26.0 g

of heptane and

35 g

of octane?

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12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 2 - Solutions>Questions>Q 44

A solution containing $30 g$ of non-volatile solute exactly in $90 g$ of water has a vapour pressure of $2.8 kPa$ at $298 K$ . Further, $18 g$ of water is then added to the solution and the new vapour pressure becomes $2.9 kPa$ at $298 K$ . Calculate molar mass of the solute.

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 2 - Solutions>Questions>Q 44

A solution containing $30 g$ of non-volatile solute exactly in $90 g$ of water has a vapour pressure of $2.8 kPa$ at $298 K$ . Further, $18 g$ of water is then added to the solution and the new vapour pressure becomes $2.9 kPa$ at $298 K$ . Calculate the vapour pressure at $298 K$ .

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 2 - Solutions>Questions>Q 45

Two elements A and B form compounds having formula

{AB}_{2}

and

{AB}_{4}

. When dissolved in

20 g

of benzene

(C_{6} H_{6})

,

1 g

of

{AB}_{2}

lowers the freezing point by

2.3 K

whereas

1.0 g

of

{AB}_{4}

lowers it by

1.3 K

. The molar depression constant for benzene is

5.1 K kg {mol}^{- 1}

. Calculate atomic masses of

A

and

B

.

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Chemistry>Chapter 2 - Solutions>Questions>Q 46

Calculate the depression in the freezing point of water when

10 g

of

{CH}_{3} {CH}_{2} CHClCOOH

is added to

250 g

of water.

K_{a} = 1.4 \times 10^{- 3}, K_{f} = 1.86 K kg {mol}^{- 1}

.

What is an ideal solution?

Important Questions on Solutions

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