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Q1.20:
A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Solution :

Step 1: Analyzing the Cane Sugar Solution

Given that the solution is 5% by mass, we interpret this as $5\text{ g}$ of solute in $95\text{ g}$ of solvent (water), totaling $100\text{ g}$ of solution.

  • Mass of cane sugar ($W_2$) = $5\text{ g}$
  • Molar mass of cane sugar ($C_{12}H_{22}O_{11}$, $M_2$) = $(12 \times 12) + (22 \times 1) + (11 \times 16) = 342\text{ g mol}^{-1}$
  • Mass of solvent ($W_1$) = $95\text{ g} = 0.095\text{ kg}$
  • Freezing point of pure water ($T_f^\circ$) = $273.15\text{ K}$
  • Freezing point of cane sugar solution ($T_f$) = $271\text{ K}$
  • Depression in freezing point ($\Delta T_f$) = $T_f^\circ - T_f = 273.15 - 271 = 2.15\text{ K}$

Step 2: Determining the Cryoscopic Constant ($K_f$)

Using the formula for depression in freezing point [Per the law of freezing point depression]:

$\Delta T_f = K_f \times m$, where $m$ is the molality ($m = \frac{W_2 \times 1000}{M_2 \times W_1}$)

$2.15 = K_f \times \frac{5 \times 1000}{342 \times 95}$

$2.15 = K_f \times \frac{5000}{32490}$

$K_f = \frac{2.15 \times 32490}{5000} = 13.97\text{ K kg mol}^{-1}$

Step 3: Analyzing the Glucose Solution

  • Mass of glucose ($W_2$) = $5\text{ g}$
  • Molar mass of glucose ($C_6H_{12}O_6$, $M_2$) = $(6 \times 12) + (12 \times 1) + (6 \times 16) = 180\text{ g mol}^{-1}$
  • Mass of solvent ($W_1$) = $95\text{ g} = 0.095\text{ kg}$
  • $K_f = 13.97\text{ K kg mol}^{-1}$ (derived from water solvent property)

Step 4: Calculating Freezing Point of Glucose Solution

Calculate molality ($m$) for glucose:

$m = \frac{5 \times 1000}{180 \times 95} = \frac{5000}{17100} = 0.2924\text{ mol kg}^{-1}$

Calculate $\Delta T_f$ for glucose:

$\Delta T_f = K_f \times m = 13.97 \times 0.2924 = 4.08\text{ K}$

Calculate final freezing point ($T_f$):

$T_f = T_f^\circ - \Delta T_f = 273.15 - 4.08 = 269.07\text{ K}$

Solute concentration (m) ΔTf (K) Cane Sugar Glucose

Final Solution: The freezing point of the 5% glucose solution is 269.07 K.


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