Question

Molality of an aqueous solution of glucose (Molar mass = 180 gmol–1) is 1.0m. 590 g of this solution is cooled and the temperature is kept at –3°C. The amount of ice separated out is nearly (kf for water = 2K kg/mol)​

Answer 1

Molality of an aqueous solution of glucose (Molar mass = 180 gmol–1) is 1.0m. 590 g of this solution is cooled and the temperature is kept at –3°C.

To find : The amount of ice seperated out.

solution : let m' be the molality of the solution after the ice seperates out at - 3°C.

using formula, ∆Tf = kf × m'

⇒m' = ∆Tf/kf = 3/2 = 1.5 molal.

let us now calculate the amount of ice seperated.

initially the Molality is 1 molal and wt of solution is 590g.

1 mol of glucose is dissolved in 1000g of H₂O.

or 180g of glucose is dissolved in 1000g of H₂O.

so (1000 + 180) = 1180g of solution contained 180g of glucose.

so 1000g of solution contained 180/1180 × 1000 = 152.54g of glucose.

amount of H₂O = 1000g - 152.54g = 847.46g

Now, when ice seperates out, the Molality is 1.5 and weight of glucose remains same as before.

because (1.5 × 180)g of glucose is present in 1000g of H₂O.

so 152.54g of glucose should be in (1000 × 152.54)/(1.5 × 180) = 564.96g of H₂O.

Thus amount of ice seperated = 847.46g - 564.96g = 282.5g