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Byju's Answer

Standard XII

Chemistry

Raoult's Law

The vapour pr...

Question

The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of the water is 0.9, the vapour pressure of the solution will be:

30 mm of Hg

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24 mm of Hg

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21 mm of Hg

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27 mm of Hg

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Solution

The correct option is B 27 mm of Hg
Given that:
$P_{A}^{o} = 30 m m o f H g, x_{H_{2} O} = 0.9$

Vapour pressure of the solution-
$P_{A} = P_{A}^{0} x_{A}$

$∴ v a p o u r p r e s s u r e$ = $30 \times 0.9 = 27$ mm of Hg
Hence, option (D) is correct.

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The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of the water is 0.9, the vapour pressure of the solution will be:

The correct option is B 27 mm of HgGiven that:PoA=30 mm of Hg,xH2O=0.9Vapour pressure of the solution-PA=P0AxA∴vapour pressure = 30×0.9=27 mm of HgHence, option (D) is correct.

The correct option is B 27 mm of Hg
Given that:
$P_{A}^{o} = 30 m m o f H g, x_{H_{2} O} = 0.9$

Vapour pressure of the solution-
$P_{A} = P_{A}^{0} x_{A}$

$∴ v a p o u r p r e s s u r e$ = $30 \times 0.9 = 27$ mm of Hg
Hence, option (D) is correct.