Find the molar mass of solute when 2 gram dissolved in 60 gram benzene at $30^{\circ} \mathrm{C}$ and…

Find the molar mass of solute when 2 gram dissolved in 60 gram benzene at $30^{\circ} \mathrm{C}$ and relative lowering of vapour pressure is 0.06 . (Molar mass of benzene is $78 \mathrm{~g} \mathrm{~mol}$ )
  1. 17.4 gram $\mathrm{mol}^{-1}$
  2. 35.2 gram mol-1
  3. 43.3 gram $\mathrm{mol}^{-1}$
  4. 24.2 gram mol-1

Solution

$\frac{\Delta \mathrm{P}}{\mathrm{P}_{\mathrm{A}}^{\circ}}=\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \times \frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{A}}}$ $0.06=\frac{2}{M_B} \times \frac{78}{60}$ $M_B=\frac{78}{30 \times 0.06}=43.3 \mathrm{~g} / \mathrm{mol}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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