Class 12th Chemistry 2017 Set1 Outside Delhi Board Paper Solution
(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 )
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]
âTf = kf x [wB / mB x wA(kg)]
âTf 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
âTf = Kf m
âTf = 342x9 / 250 X 10 x1000 / 180x90
âTf = 7.6k
Hence,
freezing point of glucose solution
= 273.15 K − 7.617K = 265.53K
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