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Q.
The mass of a non-volatile solute of molar mass $40gmol^{- 1}$ that should be dissolved in $114g$ of octane to lower its vapour pressure by $20\%$ is :
NTA AbhyasNTA Abhyas 2020
Solution:
Using Raoult's Law,
$\frac{P_{o} - P_{s}}{P_{s}}=\frac{W}{m}\times \frac{M}{W}$
If $P_{o}=100mm,$ then $P_{s}=80mm$
$\frac{100 - 80}{80}=\frac{w}{114}\times \frac{114}{40}$
$\therefore w=10g$