Question

Vapour pressure of a solution containing 18 g of glucose and 178.2 g of water at 100°C is:
(Vapour pressure of pure water at 100°C = 760 torr)

• A. 76.0 torr
• B. 752.0 torr
• C. 7.6 torr
• D. 3207.6 torr

Hint:

The calculation involves the mole fractions and vapor pressure relationship for the solute and solvent.

Answer 1

The Correct Option is B

\( n_{H_2O} = \frac{178.2}{18} = 9.9; \, n_{C_6H_{12}O_6} = \frac{18}{180} = 0.1 \)

\( \chi_{C_6H_{12}O_6} = \frac{0.1}{10} = 0.01 \)

\( \frac{P_0 - P_s}{P_0} = \chi_{C_6H_{12}O_6} \)

\( \frac{760 - P_s}{760} = 0.01 \)

\( 760 - P_s = 7.6 \)

\( P_s = 752.4 \, \text{torr} \)