Q22 / 160 · AP EAMCET 2024 (21 May Shift 1) · Chemistry

A solution was prepared by dissolving 0.1 mole of a nonvolatile solute in 0.9 moles of water. What is the relative lowering of vapour pressure of solution?

  • A
    0.9
  • B
    0.5
  • Correct answer
    C
    0.1
  • D
    0.05

Solution

Relative lowering of vapour pressure $=\frac{\Delta \mathrm{P}}{\mathrm{P}}$ $\begin{aligned} & \therefore \quad \frac{\Delta \mathrm{P}}{\mathrm{P}}=\chi_{\text {solute }} \\ & \frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solute }}+\mathrm{n}_{\text {solvent }}}=\frac{0.1}{0.1+0.9}=\frac{0.1}{1} \\ & \therefore \quad \frac{\Delta \mathrm{P}}{\mathrm{P}}=0.1 \end{aligned}$

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

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