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Raoult's Law

Vapour pressu...

Question

Vapour pressure of $C C l_{4}$ at $25^{o} C$ is $143 m m H g$ . If $0.5 g m$ of non-volatile solute (mol. mass 65) is dissolved in $100 m l C C l_{4}$ , then the vapour pressure of the solution at $25^{o} C$ is:
[Given: density of $C C l_{4} = 1.58 g / c m^{3}$ ]

141.93 m m

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94.39 m m

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199.34 m m

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143.99 m m

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Solution

The correct option is A $141.93 m m$
Weight of $C C l_{4} = d \times V = 100 \times 1.58 = 158 g = W_{2}$

$W_{1} = W e i g h t o f n o n - v o l a t i l e s o l u t e = 0.5 g$

$M_{1} = 65 g$ , $M_{2} = 154 g$

Relative lowering of vapour pressure $= \frac{P_{0} - P_{S}}{P_{0}} = \frac{W_{1} \times M_{2}}{M_{1} \times W_{2}}$

$∴ \frac{143 - P_{S}}{143} = \frac{0.5 \times 154}{65 \times 158}$

$∴ P_{S} = 141.93 m m$

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Similar questions

Vapour pressure of CCl₄ at 25⁰C is 143mm of Hg. 0.5 gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl₄. Find the vapour pressure of the solution (Density of CCl₄=1.58g/cm²)

Q. Vapour pressure of

C C l_{4}

25^{O} C

is 143mm Hg. 0.5gm of a non-volatile solute (mol. wt. 65) is dissolved in 100ml of

C C l_{4}

. Find the vapour pressure of the solution.

(Density of

C C l_{4} = 1.58 g m / {c m}^{3}

)

Q. Vapour pressure of

C C l_{4}

25^{o}

C is

143

mm of Hg.

0.5

g of a non-volatile solute (mol. Mass

= 65

) is dissolved in

100

C C l_{4}

. Find the vapour pressure of the solution: (Density of

C C l_{4} = 1.58 g / c m^{3}

)

Q. Vapour pressure of

{C C I}_{4}

24^{o} C

143 m m H g 0.05

g of the non-volatile solute (mol.wt.=65) is dissolved in

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. Find the vapour pressure of the solution (Density of

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Q. A solution containing

30 g

of a non-volatile solute in exactly

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of water has a vapour pressure of

21.85 mm of Hg

25^{o} C .

Further,

18 g

of water is then added to the solution, the new vapour pressure becomes

22.15 mm of Hg

25^{o} C .

Calculate:
(i) Molar mass of the solute (in g/mol) and
(ii) Vapour pressure of water at

25^{o} C .

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Vapour pressure of CCl4 at 25oC is 143 mmHg. If 0.5 gm of non-volatile solute (mol. mass 65) is dissolved in 100 ml CCl4, then the vapour pressure of the solution at 25oC is: [Given: density of CCl4=1.58g/cm3]

The correct option is A 141.93mmWeight of CCl4=d×V=100×1.58=158g=W2W1=Weightofnon−volatilesolute=0.5gM1=65g, M2=154gRelative lowering of vapour pressure=P0−PSP0=W1×M2M1×W2∴143−PS143=0.5×15465×158∴PS=141.93mm

Vapour pressure of $C C l_{4}$ at $25^{o} C$ is $143 m m H g$ . If $0.5 g m$ of non-volatile solute (mol. mass 65) is dissolved in $100 m l C C l_{4}$ , then the vapour pressure of the solution at $25^{o} C$ is:
[Given: density of $C C l_{4} = 1.58 g / c m^{3}$ ]