The mass of glucose that should be dissolved in $50 \mathrm{~g}$ of water in order to produce the same…

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
  1. $1 \mathrm{~g}$
  2. $3 \mathrm{~g}$
  3. $6 \mathrm{~g}$
  4. $18 \mathrm{~g}$

Solution

Lowering in vapour pressure is directly proportional to moles of solute particles $\begin{aligned} (\Delta p)_{\text {glucose }} &=(\Delta p)_{\text {urea }} \\ \left(\chi_{B}\right)_{\text {glucose }} &=\left(\chi_{B}\right)_{\text {urea }} \\ \frac{W_{B}}{50} \times \frac{18}{180} &=\frac{1 \times 18}{50 \times 60} \\ W_{B} &=3 \mathrm{~g} \end{aligned}$

Asked in: MHT CET Full Test 10

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