MEDIUM

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What weight of glucose must be dissolved in $100 g$ of water to lower the vapour pressure by $0.20 mm Hg$ ?
(Assume dilute solution is being formed)
Given: Vapour pressure of pure water is $54.2 mm Hg$ at room temperature.Molar mass of glucose is $180 g {mol}^{- 1}$

57.14% studentsanswered this correctly

Important Questions on Solutions

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

The vapour pressure of a solution of glucose is

720 mmHg

100^{o} C

. What is the mole fraction of water in this solution?

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

The vapour pressure of acetone at

20^{o} C

185 torr

. When

1.2 g

of a non-volatile substance was dissolved in

100 g

of acetone at

20^{o} C

, its vapour pressure was

183 torr

. The molar mass

(g m o l^{- 1})

of the substance is:

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

8 g

of a non-electrolyte solute is dissolved in

114 g

n -

octane to reduce its vapour pressure to

80 %,

the molar mass (in

g {mol}^{- 1}

) of the solute is [Given that molar mass of n-octane is

114 g {mol}^{- 1}

]

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

vapour pressure of water at

293 K

17535 mm Hg .

The vapour pressure of water at

293 K

containing

25 g

of glucose dissolved in

450 g

of water is

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

At room temperature, a dilute solution of urea is prepared by dissolving

0.60 g

of urea in

360 g

of water. If the vapour pressure of pure water at this temperature is

35 m m H g

, lowering of vapour pressure will be:
(molar mass of urea

= 60 g m o l^{- 1}

)

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Benzene and toluene form nearly ideal solution. At

20 ° C

, the vapour pressure of benzene is

75 torr

and that of toluene is

22 torr

. The partial vapour pressure of benzene at

20 ° C

for a solution containing

78 g

of benzene and

46 g

of toluene (in

torr

) is

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

A set of solutions is prepared using

180 g

of water as a solvent and

10 g

of different non-volatile solutes

A, B

and

C

. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of

A = 100 g {mol}^{- 1}; B = 200 g {mol}^{- 1}; C = 10, 000 g {mol}^{- 1}

]

HARD

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

104.2 ° C,

the vapour pressure of a solution of

7.5 g

of a solute in

100 g

of water is

740 mm

Hg

and

K_{b} = 0.55,

then the boiling point of the solution is ________

HARD

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

100^{o} C

the vapour pressure of a solution of 6.5 g of a non-volatile solute in 100 g water is 732 mm. If

K_{b} = 0.52

, the boiling point of this solution will be:

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container. Over time:

HARD

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

224 mL

{SO}_{2 (g)}

298 K

and

1 atm

is passed through

100 mL

0.1 M NaOH

solution. The non-volatile solute produced is dissolved in

36 g

of water. The lowering of vapour pressure of solution (assuming the solution is dilute) (

P °_{(H_{2} O)} = 24 mm

Hg

) is

x \times 10^{- 2} mm

Hg

the value of

x

is (Integer answer)

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

In 108 g of water, 18 g of a non-volatile compound is dissolved. At

100^{o} C

the vapor pressure of the solution is 750 mm Hg. Assuming that the compound does not undergo association or dissociation, the molar mass of the compound in

g m o l^{- 1}

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Vapour pressure of water at

293 K

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.

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Which of the following statement is true?

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

The vapor pressure of a dilute solution of glucose is

750 mm

of mercury at

373 K .

The mole fraction of the solute is _________

HARD

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Derive a relation between relative lowering of vapour pressure and mole fraction of the solute.

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

A $5 %$ solution of non-volatile solute in water has vapour pressure $745 mm$ of $Hg$ at $373 K$ . Calculate the molar mass of the solute.

Given vapour pressure of pure water at $373 K$ is $760 mm$ of $Hg$ .

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Relative lowering of vapour pressure of a dilute solution of glucose dissolved in

1 kg

of water is

0.002

. The molality of the solution is

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

18 g

glucose

(C_{6} H_{12} O_{6})

is added to

178.2 g

water. The vapour pressure of water (in torr) for this aqueous solution is:

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

The mass of a non-volatile, non-electrolyte solute (molar mass

= 50 g {mol}^{- 1}

) needed to be dissolved in

114 g

octane to reduce its vapour pressure by

75 %

, is:

What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is 54.2 mm Hg at room temperature.Molar mass of glucose is 180 g mol-1

Important Questions on Solutions

Learn and Prepare for any exam you want !

What weight of glucose must be dissolved in $100 g$ of water to lower the vapour pressure by $0.20 mm Hg$ ?
(Assume dilute solution is being formed)
Given: Vapour pressure of pure water is $54.2 mm Hg$ at room temperature.Molar mass of glucose is $180 g {mol}^{- 1}$