Class 12th Chemistry 2017 Set2 Outside Delhi Board Paper Solution

Question 24

(a) A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K.

Given : (Molar mass of sucrose = 342 g mol− 1 ) (Molar mass of glucose = 180 g mol− 1 )

 

Answer

10% solution (by mass) of sucrose:

wB = 10g  [Mass of solute]

mB = 342 g/mol

wS = 100 g [Mass of solution]

wA = 100-10 = 90 g [Mass of solvent]

∆Tf = kf x [wB / mB x wA(kg)]

∆Tf for sucrose solution =  273.15K - 269.15K = 4K

4 = kf x [10x1000 / 342 x 90]

kf  =  4x342x90 / 10000

kf  =  342x9 / 250

 

10% solution (by mass) of Glucose:

wB = 10g  [Mass of solute]

wS = 100 g [Mass of solution]

mB = 180 g/mol

wA = 100-10 = 90 g [Mass of solvent]

Molality (m) = 10 x1000 / 180x90 mol kg−1

∆Tf = Kf m

∆Tf  =  342x9 / 250   X  10 x1000 / 180x90

∆Tf  = 7.6k

Hence, 

freezing point of glucose solution 

=  273.15 K − 7.617K = 265.53K

 

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