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: M2=1000×Kf×W2TfW1

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 = W2 = 5g, and mass of solvent = W1 = 95 g

Molar mass of glucose (C6H12O6) = (M2) = 180 g mol-1

Δ Tf for cane sugar solution = Tf=Tf0-Tf = 273.15 K - 271 K = 215 K

Now, using the formula,

M2=1000×Kf×W2TfW1

Rearranging the formula, we get

1000Kf=M2×Tf×W1W2        ......(1)

1000Kf=M2×Tf×W1W2      .....(2)

From equations (1) and (2),

M2×Tf×W1W2=M2×Tf×W1W2

∴ 342mol-1×2.15K×95g5 g=180g mol-1×Tf×95g5 g

Tf=342mol-1×2.15K180g mol-1 = 4.085 K

∴ Freezing point of glucose solution (Tf) = Tf0-Tf

= 273.15 K - 4.085 K

= 269.065 K

Freezing point of glucose solution is 269.065 K.

Alternate method:

Formulae: 1. m = 1000W2M2W1

2. Δ Tf = Kfm

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 = 1000W2M2W1

=1000g kg-1×5g342g mol-1×95g= 0.1539 m

Now, Δ Tf for cane sugar solution = Δ Tf = Tf0-Tf = 273.15 K - 271 K = 2.15 K

From formula (ii),

Δ Tf (cane sugar) = Kf × m

Kf=Tfm

∴ Kf=2.150.1539 = 13.97 K kg mol-1

5 % glucose solution means that mass of glucose = W2 = 5 g,

and mass of solvent = W1 = 95 g

Molar mass of glucose (C6H12O6) = (M2) = 180 g mol-1

Using formula (i),

Molality of glucose solution =

m = 1000W2M2W1

1000g kg-1×5g180g mol-1×95g = 0.2924 m

From formula (ii),

Tf(glucose)=Kf×m

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

∴ Tf(glucose)=13.97×0.2924=4.085K

∴ Freezing point of glucose solution (Tf)=Tf0-Tf

= 273.15 K - 4.085 K

= 269.065 K