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 $
