Search any question & find its solution
Question:
Answered & Verified by Expert
$6 \mathrm{~g}$ of a non-volatile solute $(\mathrm{x})$ is dissolved in $100 \mathrm{~g}$ of water. The relative lowering of vapour pressure of resultant solution is 0.006 . What is the molar mass (in $\mathrm{g} \mathrm{mol}^{-1}$ ) of $\mathrm{x}$ ?
Options:
Solution:
1082 Upvotes
Verified Answer
The correct answer is:
$180$
$\begin{aligned} & \text {} \frac{\Delta \mathrm{P}}{\mathrm{P}^0}=0.006, \mathrm{~m}_1=100 \mathrm{~g}, \mathrm{~m}_2=6 \mathrm{~g}, \mathrm{M}_1=18.0 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \frac{\Delta \mathrm{P}}{\mathrm{P}^0}=\mathrm{x}_2=\frac{\mathrm{n}_2}{\mathrm{n}_1+\mathrm{n}_2} \cong \frac{\mathrm{n}_2}{\mathrm{n}_1}\left(\mathrm{n}_1 \gg \mathrm{n}_2\right) \\ & \Rightarrow \mathrm{n}_2=\frac{\Delta \mathrm{P}}{\mathrm{P}^0} \times \mathrm{n}_1=0.006 \times \frac{100.0}{18.0}=0.0334 \\ & \Rightarrow \mathrm{M}_2=\frac{\mathrm{m}_2}{\mathrm{n}_2}=\frac{6.0}{0.0334} \cong 180 \mathrm{~g} \mathrm{~mol}^{-1}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.