Answer:
Option A
Explanation:
Vapour pressure of pure solvent
= $p^{0}$=23.8 mm of Hg
Vapour pressure of solution (p0) =23.324 mm of Hg
Thus, difference in vapour pressure($\triangle$p)
=23.8-23.324=0.476mm of Hg
$\because$ $\frac{\triangle p}{p^{0}}=\frac{w_{B}}{M_{B}} \times \frac{M_{A}}{w_{A}}$ (for dilute solution)
where, MA and wA are molar mass and actual mass of solvent respectively
MB and wB are molar mass and actual mass of solute respectively;
Given;
MA=18, wA=100
wB=MB=180
$w_{B}=\frac{\triangle p \times M_{B} \times w_{A}}{p^{0} \times M_{A}}$
=$=\frac{0.02 \times180\times100}{18}$ $\left[\frac{\triangle p}{p^{0}}=\frac{0.476}{23.8}=0.02\right]$
=20.00
Hence, option (a) is the correct answer
