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 = T_{1}

Freezing point of glucose solution = 273.15 = T_{2}

Concentrations of both solutions = 10% by mass

We know that ∆T_{f} = K_{f} ×m, where

∆T_{f} = depression in freezing point = T_{2} – T_{1}

K_{f} = 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 ΔT_{f} 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) = w_{2} = 10% of 100 = 10g

molar mass of sucrose = M_{2} = 342g/mol

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

m of sucrose =

⇒ m = 0.325

now, ∆T_{f} = K_{f}m

273.15 – 269.15 = K_{f} 0.325

K_{f} = 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

K_{f} = 12.3 as obtained above

So, ∆T_{f} = K_{f} × m

= 12.3 × 0.617 = 7.59

So, freezing point of solution = 273.15 – 7.59

= 265.56K

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