What will be the molar mass of solute if vapour pressure of pure benzene is $450 \mathrm{~mm}$ $\mathrm{Hg}$…

What will be the molar mass of solute if vapour pressure of pure benzene is $450 \mathrm{~mm}$ $\mathrm{Hg}$ when $1 \cdot 5 \mathrm{~g}$ of non volatile solute is added to $30 \mathrm{~g}$ of benzene? (Vapour pressure of solution $=400 \mathrm{~mm} \mathrm{Hg}$, atomic mass $\mathrm{C}=12, \mathrm{H}=1$)
  1. $35 \cdot 1 \mathrm{~g} \mathrm{~mol}^{-1}$
  2. $26 \cdot 1 \mathrm{~g} \mathrm{~mol}^{-1}$
  3. $28 \cdot 4 \mathrm{~g} \mathrm{~mol}^{-1}$
  4. $30 \cdot 0 \mathrm{~g} \mathrm{~mol}^{-1}$

Solution

$\mathrm{P}^{0}=450 \mathrm{~mm} \mathrm{~Hg},\quad\mathrm{P}=400 \mathrm{~mm} \mathrm{~Hg}, \quad \mathrm{M}_{1}=78 \mathrm{~g} \mathrm{~mol}^{-1}$ (molar mass of benzene), $\mathrm{W}_{1}=30 \mathrm{~g}, \mathrm{~W}_{2}=1.5 \mathrm{~g}, \mathrm{M}_{2}=?$ $\mathrm{M}_{2}=\frac{\mathrm{W}_{2} \mathrm{M}_{1}}{\mathrm{~W}_{1}} \times \frac{\mathrm{P}_{0}}{\left(\mathrm{P}_{0}-\mathrm{P}\right)}$ $\therefore \mathrm{M}_{2}=\frac{1.5 \times 78}{30} \times \frac{450}{(450-400)}=35.1 \mathrm{~g} \mathrm{~mol}^{-1}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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