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The mass of glucose that should be dissolved in $50 \mathrm{~g}$ of water in order to produce the same lowering of vapour pressure as is produced by dissolving lg of urea in the same quantity of water is
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The correct answer is:
$3 \mathrm{~g}$
Lowering in vapour pressure is directly proportional to moles of solute particles
$$
\begin{aligned}
(\Delta p)_{\text {glucose }} &=(\Delta p)_{\text {urea }} \\
\left(\chi_{B}\right)_{\text {glucose }} &=\left(\chi_{B}\right)_{\text {urea }} \\
\frac{W_{B}}{50} \times \frac{18}{180} &=\frac{1 \times 18}{50 \times 60} \\
W_{B} &=3 \mathrm{~g}
\end{aligned}
$$
$$
\begin{aligned}
(\Delta p)_{\text {glucose }} &=(\Delta p)_{\text {urea }} \\
\left(\chi_{B}\right)_{\text {glucose }} &=\left(\chi_{B}\right)_{\text {urea }} \\
\frac{W_{B}}{50} \times \frac{18}{180} &=\frac{1 \times 18}{50 \times 60} \\
W_{B} &=3 \mathrm{~g}
\end{aligned}
$$
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