Molality of solution =P0−PP×1000Mw
Where P0=Vapour pressure of pure solvent=40.18 atm
P=Vapour pressure of solution
Mw=Molar mass of solvent
∴Molality (m)=40.18−4040×100018
⇒Molality=0.25 m
Now, Molality =nglucoseWwater (1kg)
0.25=W1801
⇒W=0.25×180=45 g
Now,
1000 g water contains 45 g glucose
So, 100 g solution contains
=451045×100=4.306 g glucose
⇒ weight of water = 100 g – 4.31 g = 95.69 g
Now, we know,
ΔTf=K×m
⇒0−(−0.93)=1.86×m
⇒m=0.931.86=0.5 m
Now, m=nglucoseWwater(kg)
0.5=Wglucose/1801
⇒Wg=0.5×180=90g
1000 g of water contains 90 g glucose
⇒ 90 g glucose is present in 1000 g water
∴4.31 g glucose will be present in=100090×4.31
=47.88g
∴Mass of ice =95.7−47.88 g=47.82 g