Vapour pressure in mm Hg of 0.1 mole of ure | vedclass

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(C) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute.$n_{	ext{urea}} = 0.1 \ mol$$n_{

Vedclass · Accepted Answer

$23.76$

Vedclass · Answer

(C) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute.$n_{	ext{urea}} = 0.1 \ mol$$n_{	ext{water}} = \frac{180 \ g}{18 \ g/mol} = 10 \ mol$Mole fraction of urea $(x_2)$ = $\frac{0.1}{0.1 + 10} = \frac{0.1}{10.1} \approx 0.0099$Using the formula $\frac{p^{\circ} - p_s}{p^{\circ}} = x_2$:$\frac{24 - p_s}{24} = \frac{0.1}{10.1}$$24 - p_s = 24 	imes 0.0099 = 0.2376$$p_s = 24 - 0.2376 = 23.7624 \ mm \ Hg \approx 23.76 \ mm \ Hg$

List-$I$	List-$II$
$A$. Ethane	$1$. $2$ $sp$ carbons
$B$. Ethylene	$2$. $6$ $sp^2$ carbons
$C$. Acetylene	$3$. $2$ $sp^3$ carbons
$D$. Benzene	$4$. $2$ $sp^2$ carbons
	$5$. $1$ $sp$ and $1$ $sp^2$ carbons

Vapour pressure in $mm \ Hg$ of $0.1 \ mole$ of urea in $180 \ g$ of water at $25^{\circ} C$ is (The vapour pressure of water at $25^{\circ} C$ is $24 \ mm \ Hg$)

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