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Question

Chemistry Question on Solutions

What will be the freezing point of 1%1\% solution of glucose in water, given that molal depression constant for water is 1.84Kmol1 1.84\,K\, mol^{-1}

A

272.898K272.898\, K

B

0.102C0.102\,^{\circ}C

C

273K273\, K

D

108C108\,^{\circ}C

Answer

272.898K272.898\, K

Explanation

Solution

Mass of solution =100g=100\, g
Mass of glucose =1g=1\, g
Mass of solvent =1001=99g=100-1=99\, g
w=w = mass of solute glucose,
W=W = mass of solvent m=m = molar mass of solute
ΔTf×wm×1W/1000\Delta T_{f} \times \frac{w}{m} \times \frac{1}{W / 1000}
=1.84×1180×199/1000=0.103=1.84 \times \frac{1}{180} \times \frac{1}{99 / 1000}=0.103
Freezing point of the solution =2730.103=273-0.103
=272.897K=272.897\, K