Q4 / 200 · Manipal (MET) 2012 · Chemistry

The relative lowering of vapour pressure of an aqueous solution containing a non volatile solute is 0.0125 . The molality of the solution is

  • Correct answer
    A
    0.70
  • B
    0.30
  • C
    0.125
  • D
    0.07

Solution

According to Raoult's law, "Relative lowering of vapour pressure $=$ mole fraction of solute" $\left(X_B\right)=0.0125$ $\begin{array}{rr}\because & X_B=\frac{m \cdot M_A}{10000+m \cdot M_A} \\ \therefore & 0.0125=\frac{m \times 18}{1000+m \times 18} \\ 12.5+0.225 m=18 m \\ & 17.775 m=12.5 \\ & m=\frac{12.5}{17.775}=0.70\end{array}$

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

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