Advertisement Remove all ads

A 4% Solution(W/W) of Sucrose (M = 342 G Mol−1) in Water Has a Freezing Point of 271.15 K. - Chemistry

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

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)

Advertisement Remove all ads

Solution

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

w2 (solute) = 4 g
w1 (solvent) = 96 g
M(solute) = 342 g mol−1

∆Tf = kf m
Tf = 271.15
f=27315

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

f - Tf = 0.48

Tf = T°f - 0.48

= 273.15 - 0.48

= 272.67

Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point
  Is there an error in this question or solution?
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×