A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K

Given : (Molar mass of sucrose = 342 g mol^{−1})

(Molar mass of glucose = 180 g mol^{−1})

#### Solution

Here, ΔT_{f} = (273.15 − 269.15) K = 4 K

Molar mass of sugar (C_{12}H_{22}O_{11}) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol^{−1}

10% solution (by mass) of sucrose (cane sugar) in water means 10 g of cane sugar is present in (100 − 10)g = 90 g of water.

Now, number of moles of cane sugar = `10/342 = 0.0292 mol`

Therefore, molality (m) of the solution, = `(0.0292 xx 1000)/90 = 0.3244 mol kg^(-1)`

Applying the relation,

ΔT_{f} = K_{f} × m

`=> K_f = (triangleT_t)/m = 4/0.3244 = 12.33 "K Kg mol"^(-1)`

Molar mass of glucose (C_{6}H_{12}O_{6}) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol^{−1}

10% glucose in water means 10 g of glucose is present in (100 − 10) g = 90 g of water.

∴ Number of moles of glucose = `10/180` mol

= 0.0555 mol

Therefore, molality (m) of the solution, = `(0.0555 xx 1000)/90` = 0.6166 mol kg^{−1}

Applying the relation,

ΔT_{f} = K_{f} × m

= 12.33 K kg mol^{−1} × 0.6166 mol kg^{−1}

= 7.60 K (approximately)

Hence, the freezing point of 10% glucose solution is (273.15 − 7.60) K= 265.55 K.