Question

Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol^–1) in 250 g of water. (K_f of water = 1.86 K kg mol^–1)

Answer 1

Given,

W₂ = 60 g
M₂ = 180 g mol^-1
W₁ = 250 g
K_f = ​1.86 K kg mol^–1

ΔT_f = k_fm

`T_f^0 - T_f = (K_f xx w_2 xx 1000)/M_2xxw_1`

`2.73 - T_f = (1.86 xx 60 xx 1000)/(180xx250)`

`2.73 - T_f = (1.86 xx 60 xx 1000)/(180 xx 250)`

273.15 - T_f = 2.48

T_f = 273.15 - 2.48 = 270.67 K

Hence, the freezing point of water is 270.67 K or -2.48°C