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$6 \mathrm{~g}$ of urea is dissolved in $90 \mathrm{~g}$ of water. Relative lowering of vapour pressure is
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Verified Answer
The correct answer is:
$0.02$
$\frac{p^{\circ}-p_{\mathrm{S}}}{p^{\circ}}=\frac{w_{2} M_{1}}{M_{2} w_{1}}$
Where, $w_{2}$ and $M_{2}$ are weight and molecular mass of solute.
$M_{1}$ and $w_{1}$ are molecular mass and weight of solvent.
$$
\frac{\Delta p}{p^{\circ}}=\frac{6}{60} \times \frac{18}{90}=\frac{1}{50}=0.02
$$
Where, $w_{2}$ and $M_{2}$ are weight and molecular mass of solute.
$M_{1}$ and $w_{1}$ are molecular mass and weight of solvent.
$$
\frac{\Delta p}{p^{\circ}}=\frac{6}{60} \times \frac{18}{90}=\frac{1}{50}=0.02
$$
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