A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has freezing point of 271 K. Calculate the freezing point of 5% aqueous glucose solution. - Chemistry

Exercise | Q 10 | Page 46

A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has a freezing point of 271 K. Calculate the freezing point of 5% aqueous glucose solution.

Solution:

Given: Percentage by mass of cane sugar solution = 5 %
Percentage by mass of glucose solution = 5 %,
Freezing point of cane sugar solution = 271 K
Molar mass of cane sugar = 342 g mol^-1

To find: Freezing point of glucose solution

Formula: ${\displaystyle {\text{M}}_2=\frac{1000\times {\text{K}}_{\text{f}}\times {\text{W}}_2}{△{\text{T}}_{\text{f}}{\text{W}}_1}}$

Calculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W₂) = 5 g, and mass of solvent (W₁) = 95 g.

5 % glucose solution means that mass of glucose = ${\displaystyle {\text{W}}_2^{\prime }}$ = 5g, and mass of solvent = ${\displaystyle {\text{W}}_1^{\prime }}$ = 95 g

Molar mass of glucose (C₆H₁₂O₆) = ${\displaystyle \left({\text{M}}_2^{\prime }\right)}$ = 180 g mol^-1

Δ T_f for cane sugar solution = ${\displaystyle △{\text{T}}_{\text{f}}={\text{T}}_{\text{f}}^0-{\text{T}}_{\text{f}}}$ = 273.15 K - 271 K = 215 K

Now, using the formula,

${\displaystyle {\text{M}}_2=\frac{1000\times {\text{K}}_{\text{f}}\times {\text{W}}_2}{△{\text{T}}_{\text{f}}{\text{W}}_1}}$

Rearranging the formula, we get

${\displaystyle 1000{\text{K}}_{\text{f}}=\frac{{\text{M}}_2\times △{\text{T}}_{\text{f}}\times {\text{W}}_1}{{\text{W}}_2}}$        ......(1)

${\displaystyle 1000{\text{K}}_{\text{f}}=\frac{{\text{M}}_2^{\prime }\times △{\text{T}}_{\text{f}}^{\prime }\times {\text{W}}_1^{\prime }}{{\text{W}}_2^{\prime }}}$      .....(2)

From equations (1) and (2),

${\displaystyle \frac{{\text{M}}_2\times △{\text{T}}_{\text{f}}\times {\text{W}}_1}{{\text{W}}_2}=\frac{{\text{M}}_2^{\prime }\times △{\text{T}}_{\text{f}}^{\prime }\times {\text{W}}_1^{\prime }}{{\text{W}}_2^{\prime }}}$

∴ ${\displaystyle \frac{342{\text{mol}}^{-1}\times 2.15\text{K}\times 95\text{g}}{\text{5 g}}=\frac{180{\text{g mol}}^{-1}\times △{\text{T}}_{\text{f}}^{\prime }\times 95\text{g}}{\text{5 g}}}$

${\displaystyle △{\text{T}}_{\text{f}}^{\prime }=\frac{342{\text{mol}}^{-1}\times 2.15\text{K}}{180{\text{g mol}}^{-1}}}$ = 4.085 K

∴ Freezing point of glucose solution (T_f) = ${\displaystyle {\text{T}}_{\text{f}}^0-△{\text{T}}_{\text{f}}^{\prime }}$

= 273.15 K - 4.085 K

= 269.065 K

Freezing point of glucose solution is 269.065 K.

Alternate method:

Formulae: 1. m = ${\displaystyle \frac{1000{\text{W}}_2}{{\text{M}}_2{\text{W}}_1}}$

2. Δ T_f = K_fm

Calculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W₂) = 5 g, and mass of solvent (W₁) = 95 g.

Now, using formula (i),

Molality of cane sugar solution = m = ${\displaystyle \frac{1000{\text{W}}_2}{{\text{M}}_2{\text{W}}_1}}$

${\displaystyle =\frac{1000{\text{g kg}}^{-1}\times 5\text{g}}{342{\text{g mol}}^{-1}\times 95\text{g}}}$= 0.1539 m

Now, Δ T_f for cane sugar solution = Δ T_f = ${\displaystyle △{\text{T}}_{\text{f}}^0-{\text{T}}_{\text{f}}}$ = 273.15 K - 271 K = 2.15 K

From formula (ii),

Δ T_f (cane sugar) = K_f × m

${\displaystyle {\text{K}}_{\text{f}}=\frac{△{\text{T}}_{\text{f}}}{\text{m}}}$

∴ ${\displaystyle {\text{K}}_{\text{f}}=\frac{2.15}{0.1539}}$ = 13.97 K kg mol^-1

5 % glucose solution means that mass of glucose = ${\displaystyle {\text{W}}_2^{\prime }}$ = 5 g,

and mass of solvent = ${\displaystyle {\text{W}}_1^{\prime }}$ = 95 g

Molar mass of glucose (C₆H₁₂O₆) = ${\displaystyle \left({\text{M}}_2^{\prime }\right)}$ = 180 g mol^-1

Using formula (i),

Molality of glucose solution =

m = ${\displaystyle \frac{1000{\text{W}}_2}{{\text{M}}_2{\text{W}}_1}}$

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

From formula (ii),

${\displaystyle △{\text{T}}_{\text{f}}^{\prime }\left(\text{glucose}\right)={\text{K}}_{\text{f}}\times \text{m}}$

(⸪ Since solvent is same, Kf is same for cane sugar and glucose solutions.)

∴ ${\displaystyle △{\text{T}}_{\text{f}}^{\prime }\left(\text{glucose}\right)=13.97\times 0.2924=4.085\text{K}}$

∴ Freezing point of glucose solution ${\displaystyle \left({\text{T}}_{\text{f}}\right)={\text{T}}_{\text{f}}^0-△{\text{T}}_{\text{f}}^{\prime }}$

= 273.15 K - 4.085 K

= 269.065 K