The total vapour pressure of a 4 mole $\%$ solution of $\mathrm{NH}_{3}$ in water at $293 \mathrm{~K}$ is…

The total vapour pressure of a 4 mole $\%$ solution of $\mathrm{NH}_{3}$ in water at $293 \mathrm{~K}$ is $50.0$ torr. The vapour pressure of pure water is $17.0$ torr at this temperature. Applying Henry's and Raoult's laws, the total vapour pressure for a 5 mole $\%$ solution is
  1. $58.25$ torr
  2. 33 torr
  3. $42.1$ torr
  4. $52.25$ torr

Solution

The given data are $\mathrm{P}_{\text {water }}=17.0$ torr
$\mathrm{P}_{\text {total }}(4$ mole $\%$ solution $)$ $=\mathrm{P}_{\mathrm{NH}_{3}}+\mathrm{P}_{\text {water }}=50.0$ torr
$\mathrm{X}_{\mathrm{NH}_{3}}=0.04$ and $\mathrm{X}_{\text {water }}=0.96$
Now according to Raoult's law;
$\mathrm{P}_{\text {water }}=\mathrm{X}_{\text {water }} \mathrm{P}_{\text {water }}^{\circ}$
$=0.96 \times 17.0$ torr $=16.32$ torr
Now Henry's law constant for ammonia is
$\mathrm{K}_{\mathrm{H}}\left(\mathrm{NH}_{3}ight)=\frac{\mathrm{P}_{\mathrm{NH}_{3}}}{\mathrm{X}_{\mathrm{NH}_{3}}}=\frac{33.68 \text { torr }}{0.04}=842$ torr
Hence, for 5 mole $\%$ solution, we have
$\mathrm{P}_{\mathrm{NH}_{3}}=\mathrm{K}_{\mathrm{H}}\left(\mathrm{NH}_{3}ight) \mathrm{X}_{\mathrm{NH}_{3}}$
$=(842$ torr $)(0.05)=16.15$ torr
Thus, $\mathrm{P}_{\text {total }}(5$ mole $\%$ solution)
$=\mathrm{P}_{\mathrm{NH}_{3}}+\mathrm{P}_{\text {water }}=42.1+16.15=58.25$ torr .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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