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प्रश्न
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)
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उत्तर
4% solution(w/w) of sucrose ⇒ 4 g sucrose in 96 g water
w2 (solute) = 4 g
w1 (solvent) = 96 g
M2 (solute) = 342 g mol−1
∆Tf = kf m
Tf = 271.15
T°f=273⋅15
ΔTf = Tf - 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`
∆Tf = kf m.
* 5% solution(w/w) of glucose in water ⇒ 5 g glucose in 95 g H2O
w2 = 5 g
w1 = 95 g
M2 = 180 g
∆Tf = ` 16.39 xx (5 xx 100)/(95 xx 180) = 0.479`
∆Tf ≅ 0.48
T°f - Tf = 0.48
Tf = T°f - 0.48
= 273.15 - 0.48
= 272.67
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