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Standard XII

Chemistry

Raoult's Law

Vapour pressu...

Question

Vapour pressure of $C C l_{4}$ at $25^{o}$ C is $143$ mm of Hg. $0.5$ g of a non-volatile solute (mol. Mass $= 65$ ) is dissolved in $100$ mL $C C l_{4}$ . Find the vapour pressure of the solution: (Density of $C C l_{4} = 1.58 g / c m^{3}$ )

141.93

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94.39

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199.34

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143.99

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Solution

The correct option is A $141.93$ mm
$P_{0} = 143$
$V = 100 m l$

$\frac{P_{0} - P}{P_{0}} = \frac{n_{B}}{n_{A}}$

$n_{B} = \frac{0.5}{65}$

$m a s s = V \times d = 100 \times 1.58 = 158$

$n_{A} = \frac{158}{154}$

$\frac{P_{0} - P}{P_{0}} = \frac{0.5}{65} \times \frac{154}{158} = 0.007$

$143 - P = .007 \times 143 = 1.07$

$\Rightarrow P = 143 - 1.07 = 141.93$

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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

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}

]

Q. 0.5 g of a nonvolatile organic compound (molecular weight 65) is dissolved in

100 m L o f C C l_{4}

. If the vapour pressure of pure

C C l_{4} i s 140 m m H g

, then calculate the vapour pressure of the solution.

(Density of C C l_{4} solution is 1.6 g m L^{- 1}

5.8 g

of non volatile solute was dissolved in 100 g of carbon disulphide

(C S_{2})

. The vapour pressure of the solution was found to be 190 mm. of Hg. Calculate the molar mass of the solute given the vapour pressure of pure

C S_{2}

is 195 mm. of Hg [Molar mass of

C S_{2} = 76 g m o l^{- 1}

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Vapour pressure of CCl4 at 25oC is 143 mm of Hg. 0.5 g of a non-volatile solute (mol. Mass =65) is dissolved in 100 mL CCl4. Find the vapour pressure of the solution: (Density of CCl4=1.58g/cm3)

The correct option is A 141.93 mmP0=143V=100mlP0−PP0=nBnAnB=0.565mass=V×d=100×1.58=158nA=158154P0−PP0=0.565×154158=0.007143−P=.007×143=1.07⇒P=143−1.07=141.93

Vapour pressure of $C C l_{4}$ at $25^{o}$ C is $143$ mm of Hg. $0.5$ g of a non-volatile solute (mol. Mass $= 65$ ) is dissolved in $100$ mL $C C l_{4}$ . Find the vapour pressure of the solution: (Density of $C C l_{4} = 1.58 g / c m^{3}$ )