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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 $180 g {mol}^{- 1}$

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Important Questions on Solutions

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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

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

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:

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Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

38.4 g

of unknown substance (molar mass

384 g {mol}^{- 1}

) and

116 g

of acetone is used to prepare a solution at

313 K

. If vapour pressure of pure acetone (molar mass

58 g {mol}^{- 1}

) is

0.842

atmosphere, what is the vapour pressure of solution ?

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Consider the following liquid - vapour equilibrium.

L i q u i d ⇌ V a p o u r

Which of the following relations is correct?

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}

]

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

9 gram anhydrous oxalic acid (mol. Wt. = 90) was dissolved in 9.9 moles of water. If vapour pressure of pure water is

P_{1}^{o}

the vapour pressure of solution is

EASY

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Which of the following statement is true?

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:

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

For a solution formed by mixing liquids

L

and

M

, the vapour pressure of

L

plotted against the mole fraction of

M

in solution is shown in the following figure, Here,

X_{L}

and

X_{M}

represent mole fractions of

L

and

M

, respectively, in the solution. The correct statement(s) applicable to this system is(are) -

Question Image

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?

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:

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

How many grams of glucose are required to prepare an aqueous solution of glucose having a vapor pressure of

23.324 mm Hg

25 ° C

100 g

of water? The vapour pressure of pure water at

25 ° C

23.8 mm Hg .

MEDIUM

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.

HARD

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

The plot of total vapour pressure as a function of mole fraction of the components of an ideal solution formed by mixing liquids

X

and

Y

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

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

MEDIUM

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Liquid

M

and liquid

N

form an ideal solution. The vapour pressures of pure liquids

M

and

N

are

450

and

700 m m H g,

respectively, at the same temperature. Then correct statements is:
(

x_{M} = M o l e f r a c t i o n o f {' M}^{'} i n s o l u t i o n;

x_{N} = M o l e f r a c t i o n o f {' N}^{'} i n s o l u t i o n;

y_{M} = M o l e f r a c t i o n o f {' M}^{'} i n v a p o u r p h a s e;

y_{N} = M o l e f r a c t i o n o f {' N}^{'} i n v a p o u r p h a s e;

)

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:

HARD

Chemistry>Physical Chemistry>Solutions>Lowering of Vapour Pressure

Liquids A and B form ideal solution over the entire range of composition. At temperature

T

, equimolar binary solution of liquids A and B has vapour pressure

45 T o r r

. At the same temperature, a new solution of A and B having mole fractions

x_{A} and x_{B}

, respectively, has vapour pressure of

22.5 T o r r

. The value of

\frac{x_{A}}{x_{B}}

, in the new solution is ____.

(Given that the vapour pressure of pure liquid A is

20 Torr

at temperature

T

)

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 $180 g {mol}^{- 1}$