Q. The relative lowering of vapour pressure of a dilute aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is about
A) 0.70
B) 0.50
C) 0.90
D) 0.80
Explanation
(a)
Relative lowering of vapour pressure = mole fraction of solute (Raoult’s law )
$\frac{p - p_{s}}{p} = x_{2}\ \$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{p - p_{s}}{p} = \frac{wM}{mW}\ \$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0125$= $\frac{wM}{mW}$
Or $\frac{w}{mW} = \frac{0.0125}{18} = 0.00070$
Hence, molality =$\frac{w}{mW} \times 1000$
$= 0.0007 \times 1000 = 0.70$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.