Solutions Question 45

Question: A solution containing 30 gms of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mm Hg at $ 25^{o}C $ . Further 18 gms of water is then added to the solution. The resulting solution has a vapour pressure of 22.15 mm Hg at $ 25^{o}C $ . Calculate the molecular weight of the solute [UPSEAT 2001]

Options:

A) 74.2

B) 75.6

C) 67.83

D) 78.7

Show Answer

Answer:

Correct Answer: C

Solution:

We have, $ \frac{p^{0}-21.85}{21.85}=\frac{30\times 18}{90\times m} $ , for I case …..(i) wt. of solvent $ =90+18=108gm $

$ \frac{p^{0}-22.15}{22.15}=\frac{30\times 18}{108\times m} $ , for II case …..(ii) By eq. (1) $ p_m^{0}-21.85m=21.85\times 6=131.1 $ By eq. (2) $ p_m^{0}-22.15m=22.15\times 5=110.75 $ 0.30m = 20.35 $ m=\frac{20.35}{0.30}=67.83 $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक