Discussion
The vapour pressure of pure benzene (molar mass=78 g mol−1 ) at a certain temperature is 0.85 bar. When 0.5 g of a non-volatile, non-electrolyte is added to 39 g of benzene, the vapour pressure was found to be 0.845 bar at the same temperature. What is the molar mass of the substance?
Answer: C
The molar mass is 170 g mol−1.
For a dilute solution, the relative lowering of vapour pressure equals the ratio of moles of solute to solvent:
\[ \begin{aligned} \frac{0.85 - 0.845}{0.85} &= \frac{0.5/M}{39/78} \\ \frac{0.005}{0.85} &= \frac{1}{M} \Rightarrow M = 170 \end{aligned} \]
The other molar masses do not satisfy this relation; 85 g mol−1, for example, is what you get by taking the moles of benzene as 1 instead of \(\frac{39}{78} = 0.5\).
Strictly, Raoult's law gives \(\frac{n_{B}}{n_{A} + n_{B}}\), but for a dilute solution \(n_{B}\) is negligible beside \(n_{A}\), which is the approximation used here.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development