HARD

JEE Main

IMPORTANT

Earn 100

A solution contains $30 g$ of a non-volatile solute in exactly $90 g$ of water, its vapour pressure is $21.85 mm$ of $Hg$ at $25 ° C .$ Further, $18 g$ of water is then added to the solution, the new vapour pressure becomes $22.15 mm$ of $Hg$ at $25 ° C$ . Calculate the vapour pressure of water at $25 ° C$ .

Important Questions on Solutions

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 2 - Solutions>DPP 2.3>Q 5

What mass of the non-volatile solute urea

({NH}_{2} {CONH}_{2})

needs to be dissolved in

100 g

of water, in order to decrease the vapour pressure of water by

25 %

? What will be the molality of the solution?

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 2 - Solutions>DPP 2.3>Q 6

Molal elevation constant of a liquid is

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 2 - Solutions>DPP 2.3>Q 7

The vapour pressure of a solution of a non-volatile electrolyte

(A)

in a solvent

(B)

95 %

of the vapour pressure of the solvent at the same temperature. If

M_{B} = 0.3 M_{A}

where

M_{B}

and

M_{A}

are molar masses of

B

and

A

respectively, the mass ratio of the solvent and solute is

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 2 - Solutions>DPP 2.3>Q 8

Elevation in the boiling point of an aqueous glucose solution is

0.6; K_{b}

for water is

0.52 K

{molality}^{- 1}

. The mole fraction of glucose in the solution is

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 2 - Solutions>DPP 2.3>Q 9

Normal boiling point

(T_{N})

is defined as the temperature when the vapour pressure of a liquid becomes equal to

1 atm

and the standard boiling point

(T_{S})

is defined as the temperature at which the vapour pressure of a liquid becomes equal to

1 bar

. Which one is not correct with regard to water?

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 2 - Solutions>DPP 2.3>Q 10

Y gm

of a non-volatile solute of molar mass

M

is dissolved in

250 g

of benzene. If

K_{b}

is molal elevation constant, then the value of

ΔT

is given by

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 2 - Solutions>DPP 2.3>Q 11

The vapour pressure of pure liquid

A

10 torr

and at the same temperature when

1 g

of solute

B

is dissolved in

20 g

A

, its vapour pressure is reduced to

9.0 torr

. If the molecular mass of

A

200 amu

, then the molecular mass of

B

is:

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 2 - Solutions>DPP 2.3>Q 12

Calculate the ebullioscopic constant for water. The heat of vaporization is

40.685 kJ {mol}^{- 1}

A solution contains 30 g of a non-volatile solute in exactly 90 g of water, its vapour pressure is 21.85 mm of Hg at 25°C. Further, 18 g of water is then added to the solution, the new vapour pressure becomes 22.15 mm of Hg at 25°C. Calculate the vapour pressure of water at 25°C.

Important Questions on Solutions

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