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.

• A. 18.0
• B. 342
• C. 60
• D. 180

Answer 1

Answer: (D)

180

Answer 2

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

$m = 180$