Q. 114.5( 8 Votes )

A 10% solution (b

Answer :

Given : (Molar mass of sucrose = 342 g mol–1)


(Molar mass of glucose = 180 g mol–1)


Freezing point of sucrose solution = 269.15K = T1


Freezing point of glucose solution = 273.15 = T2


Concentrations of both solutions = 10% by mass


We know that ∆Tf = Kf ×m, where


∆Tf = depression in freezing point = T2 – T1


Kf = cryoscopic constant


We know that


If w2 is the grams of the solute having molar mass as M2, present in w1 gram of solvent, produces the depression in freezing point ΔTf of the solvent then molality of the solute m is given by the equation-


m =


assume 100g of solution containing 90% solvent and 10% solute


mass of solute (sucrose) = w2 = 10% of 100 = 10g


molar mass of sucrose = M2 = 342g/mol


mass of solvent = w1 = 90% of 100 = 90g


m of sucrose =


m = 0.325


now, Tf = Kfm


273.15 – 269.15 = Kf 0.325


Kf = 12.3


In case of glucose, assuming 100g of solution


Mass of solute = 10% of 100 = 10g


Mass of solvent = 90g


Now, m of glucose = = 0.617


Kf = 12.3 as obtained above


So, ∆Tf = Kf × m


= 12.3 × 0.617 = 7.59


So, freezing point of solution = 273.15 – 7.59


= 265.56K


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