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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 - Chemistry

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)

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Solution

Here, ΔTf = (273.15 − 269.15) K = 4 K

Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol−1

10% solution (by mass) of sucrose (cane sugar) in water means 10 g of cane sugar is present in (100 − 10)g = 90 g of water.

Now, number of moles of cane sugar = `10/342 = 0.0292 mol`

Therefore, molality (m) of the solution, = `(0.0292 xx 1000)/90 = 0.3244 mol kg^(-1)`

Applying the relation,

ΔTf = Kf × m

`=> K_f = (triangleT_t)/m = 4/0.3244 = 12.33 "K Kg mol"^(-1)`

Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol−1

10% glucose in water means 10 g of glucose is present in (100 − 10) g = 90 g of water.

∴ Number of moles of glucose = `10/180` mol

= 0.0555 mol

Therefore, molality (m) of the solution, = `(0.0555 xx 1000)/90` = 0.6166 mol kg−1

Applying the relation,

ΔTf = Kf × m

= 12.33 K kg mol−1 × 0.6166 mol kg−1

= 7.60 K (approximately)

Hence, the freezing point of 10% glucose solution is (273.15 − 7.60) K= 265.55 K.

Concept: Carbohydrates - Preparation of Glucose from Sucrose (Cane Sugar) and Starch
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