A dilute aqueous solution of glucose shows a vapour pressure of 750 mm of Hg at 373 K. The molality of the solution is

Options

A. 13.32
B. 0.013
C. 1.35
D. 0.74

Answer

D

Explanation

`(P^@-P_s)/(P^@) = X_2` (Raoult.s law)
`(760 - 750)/760 = 0.0132`
Mole fraction of solvent `(X_1)`= 1 – 0.0132 = 0.9868 It means 0.0132 mole of glucose is present per 0.9868 mole of water.
Weight of solvent `= 0.9868 xx 18 = 17.76 g`
(molecular weight of `H_2O = 18`)
Molality `= (0.0132 xx 1000)/(17.76) = 0.74`