Q37 / 160 · AP EAMCET 2014 · Chemistry

Vapour pressure in $\mathrm{mm} \mathrm{Hg}$ of 0.1 mole of urea in $180 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C}$ is (The vapour pressure of water at $25^{\circ} \mathrm{C}$ is $24 \mathrm{~mm} \mathrm{Hg}$ )

  • A
    $2.376$
  • B
    $20.76$
  • Correct answer
    C
    $23.76$
  • D
    $24.76$

Solution

From Raoult's law for very dilute solution. $$ \begin{aligned} \frac{p^{\circ}-p_s}{p^{\circ}} & =\frac{w}{m} \times \frac{M}{W} \\ \frac{24-p_s}{24} & =0.1 \times \frac{18}{180} \\ 24-p_s & =0.24 \\ \therefore \quad p_s= & 24-0.24=23.76 \mathrm{~mm} \mathrm{Hg} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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