Solutions Question 108

Question: The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be [DPMT 2001]

Options:

A) 18.0

B) 342

C) 60

D) 180

Show Answer

Answer:

Correct Answer: D

Solution:

$ \frac{P^{0}-P_{s}}{P^{0}}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}} $ or $ 0.00713=\frac{{71.5}/{m};}{\frac{71.5}{m}+\frac{1000}{18}} $

$ m=180 $



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