Question

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

Answer 1

Here, ΔT_f = (273.15 − 271) K

= 2.15 K

Molar mass of sugar (C₁₂H₂₂O₁₁) = 12 × 12 + 22 × 1 + 11 × 16

= 342 g mol⁻¹

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g = 95 g of water.

Now, number of moles of cane sugar = `5/342` mol

= 0.0146 mol

∴ Molality of the solution (m) = `(0.0146  "mol")/(0.095  kg)`

= 0.1537 mol kg⁻¹

Applying the relation,

ΔT_f = K_f × m

⇒ K_f = `(Delta T_f)/m`

= `(2.15  K)/(0.1537  "mol"  kg^(-1))`

= 13.99 K kg mol⁻¹

Molar of glucose (C₆H₁₂O₆) = 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol⁻¹

5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.

Number of moles of glucose = `5/180` mol

= 0.0278 mol

∴ Molality of the solution (m) = `(0.0278  "mol")/(0.095  kg)`

= 0.2926 mol kg⁻¹

Applying the relation,

ΔT_f = K_f × m

= 13.99 K kg mol⁻¹ × 0.2926 mol kg⁻¹

= 4.09 K (approximately)

Freezing point of the 5% glucose solution is = 273.15 − 4.09 K

= 269.06 K

Answer 2

For a cane sugar solution,

M' = `(1000 xx K_f xx w)/(W xx Delta T_f)`

⇒ 342 = `(1000 xx K_f xx 5)/(95 xx (273.15 - 271))`

342 = `(5 xx 1000 xx K_f)/(95 xx 2.15)`    ...(i)

For glucose solution,

180 = `(1000 xx K_f xx 5)/(95 xx Delta T_f)`

180 = `(5 xx 1000 xx K_f)/(95 xx Delta T_f)`    ...(ii)

Dividing eq. (i) by eq. (ii), we have.

`342/180 = (Delta T_f)/2.15`

⇒ ΔT_f = `(342 xx 2.15)/180`

= 4.085 K

Freezing point of glucose solution = 273.15 − 4.085

= 269.07 K

∴ The freezing point of a glucose solution is 269.07 K.