Question

A 5% solution(w/W) of cane sugar (molar mass =

$342gmol^{- 1}$) has freezing point 271 K what will be the freezing point of 5% glucose (molar mass = $18gmol^{- 1}$) in water freezing point of pure water is $273.15K?$

• A. $273.07K$
• B. $269.07K$
• C. $273.15K$
• D. $260.09K$

Answer 1

Answer: (B)

$269.07K$

Answer 2

$\Delta T_{f} = \frac{k_{f} \times W_{B}}{M_{B} \times W_{A}}$

For cane sugar solution $2.15K = \frac{k_{f} \times 5}{342 \times 0.095}$

(95 g of water =$0.095kg$)

For glucose solution $\Delta T_{f} = \frac{K_{f} \times 5}{180 \times 0.095}$

$\frac{\Delta T_{f}}{2.15} = \frac{k_{f} \times 5}{180 \times 0.095} \times \frac{342 \times 0.095}{k_{f} \times 5}$

$\Delta T_{f} = \frac{342}{180} \times 2.15 = 4.085K$

Freezing point of glucose solution $= 273.15 - 4.085$

$= 269.07K$