A 4% solution(w/w) of sucrose (M = 342 g mol^{−1}) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol^{−1})^{ }in water.

(Given: Freezing point of pure water = 273.15 K)

#### Solution

4% solution(w/w) of sucrose ⇒ 4 g sucrose in 96 g water

w_{2} (solute) = 4 g

w_{1} (solvent) = 96 g

M_{2 }(solute) = 342 g mol^{−1}

∆T_{f} = k_{f} m

T_{f} = 271.15

T°_{f}=273⋅15

ΔT_{f} = T_{f }- T°_{f} = (273. 15 - 271 . 15)K = 2.0K

`k_f = (Δ"T"_f)/(m)`

`m = (w^2)/("M" xx w_1) xx 1000 = (4 xx 1000)/(96 xx 342) = 0 .122"m"`

`k_f = (2)/(0.122) = 16.39"Km"^-1`

∆T_{f} = k_{f} m.

* 5% solution(w/w) of glucose in water ⇒ 5 g glucose in 95 g H_{2}O

w_{2} = 5 g

w_{1} = 95 g

M_{2} = 180 g

∆T_{f} = ` 16.39 xx (5 xx 100)/(95 xx 180) = 0.479`

∆T_{f} ≅ 0.48

T°_{f} - Tf = 0.48

T_{f} = T°_{f} - 0.48

= 273.15 - 0.48

= 272.67