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Question
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water. (Kf of water = 1.86 K kg mol–1)
Solution
Given,
W2 = 60 g
M2 = 180 g mol-1
W1 = 250 g
Kf = 1.86 K kg mol–1
ΔTf = kfm
`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 - Tf = 2.48
Tf = 273.15 - 2.48 = 270.67 K
Hence, the freezing point of water is 270.67 K or -2.48°C
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