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The vapour pressure of water at $20 ^{\circ}C$ is 17.5 mm $Hg$ If 18 g of glucose $\left ( C_{6}H_{12}O_{6} \right )$ is added to 178.2 g of water at $20 ^{\circ}C$ the vapour pressure of the resulting solution will be

Option 1)
$17.325\: mm \: Hg$

Option 2)
$17.675\: mm \: Hg$

Option 3)
$15.750\: mm \: Hg$

Option 4)
$16.500\: mm \: Hg$

Answers (1)

best_answer

As we learnt in

Expression of relative lowering of vapour pressure -

$\frac{\Delta P}{ P^{0}}= x_{solute}$

$x_{solute}= \frac{ n_{solute}}{ n_{solute}+n_{solvent}}$

- wherein

$\Delta P \: is \: lowering \: o\! f \: v.p.$

$P^{0}\rightarrow \: vapour\: pressure\: of \: pure\: solvent$

$x_{solute}\rightarrow \: mole\: fraction \: of \:non\: volatile\: solute$

$p_{solution}= p_{A}^{0}\times \chi_{A}$

$\chi_{A}=\frac{n_{A}}{n_{B}+n_{A}}=\frac{\frac{178.2}{18}}{\frac{178}{18}+\frac{18}{180}}$

$\chi_{A}=\frac{9.9}{10}=0.99$

$p_{solution}=17.5\times 0.99$

$p_{solution}=17.325$

Correct option is 1.

Option 1)

$17.325\: mm \: Hg$

This is the correct option.

Option 2)

$17.675\: mm \: Hg$

This is an incorrect option.

Option 3)

$15.750\: mm \: Hg$

This is an incorrect option.

Option 4)

$16.500\: mm \: Hg$

This is an incorrect option.

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