What is vapour pressure of solution containing $1.8 \mathrm{~g}$ glucose in $16.2 \mathrm{~g}$ water?…

What is vapour pressure of solution containing $1.8 \mathrm{~g}$ glucose in $16.2 \mathrm{~g}$ water? $\left(\mathrm{P}_1^0=24 \mathrm{~mm} \mathrm{Hg} \text { and Molar mas of gloucose }=180 \mathrm{~g} \mathrm{~mol}^{-1}\right)$
  1. $18.1 \mathrm{~mm} \mathrm{Hg}$
  2. $15.7 \mathrm{~mm} \mathrm{Hg}$
  3. $12.4 \mathrm{~mm} \mathrm{Hg}$
  4. $23.8 \mathrm{~mm} \mathrm{Hg}$

Solution

$\begin{aligned} & \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}^0}=\mathrm{X}_{\text {solute }}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solute }}+\mathrm{n}_{\text {solvent }}} \\ & \mathrm{n}_{\text {glucose }}=\frac{1.8}{180}=0.01 \\ & \mathrm{n}_{\text {water }}=\frac{16.2}{18}=0.9 \\ & \frac{24-\mathrm{P}_{\mathrm{s}}}{24}=\frac{0.01}{0.01+0.9}=\frac{1}{91} \\ & \left(24-\mathrm{P}_{\mathrm{s}}\right) 91=24 \\ & \mathrm{P}_{\mathrm{s}}=24-\frac{24}{91}=23.8 \mathrm{~mm} \mathrm{Hg}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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