Question

In an experiment, 18.0 g of mannitol was dissolved in 100 g of water. The vapour pressure of solution at ${20}^{o}C$ was 17.226 mm of mercury. Calculate the molecular mass of mannitol. The vapour pressure of water at ${20}^{o}C$ is 17.535 mm of mercury.

• A. Molecular mass of mannitol = 171.02
• B. Molecular mass of mannitol = 181.02
• C. Molecular mass of mannitol = 161.02
• D. Molecular mass of mannitol = 121.02

Answer 1

The correct option is B Molecular mass of mannitol = 181.02

we have,

$\frac{P_B^0-P_t}{P_B^0}=x_A$

$\frac{17.535-17.226}{17.535}=\frac{x}{x+5.55}$

we get x = 0.9811

and molecular weight = 183.02 g