Question

The vapour pressure of an aqueous solution of glucose is 750 mm of Hg at 373 k . Calculate the molality and mole fraction of solute.

Answer 1

At 373K V.P of water is 750 mm Hg

We know that, (P0-P) / P0 =  w x M / m x W

molality = (P0-P) / P0 X 1000/M 

=> (760-750) / 760 X 1000/18

=> 0.73 m

Answer>> Mole fraction = (P0-P) / P0  = 10/760 = 0.0131  

Answer 2

Answer:

molality of given solution is 0.73m

mole fraction of glucose is=0.013M

Explanation:

Given is,

vapour pressure of glucose = 750 mm

at 373

Molality=?

Mole fraction =?

solution:

The Vapour pressure of glucose

is 760 mm Hg at 373 k We

know

(P_{0} - P)/P_{0} = (wM)/(mW)

p molaiily= P 0 -P P 0 *1000/M

= 760-750 760*100918 =0.73m

mole fraction = P o -P P o = 10 70 0

=0.013 I