Question

Depression of freezing point of 0.01 molal aqueous solution ch3cooh is 0.0204°c .1 molal urea solution at-1.86 assuming m=m ph?

Answer 1

Assume you have 1.00 L (1000 mL) of solution.

d = m / V

m = d x V = 1.23 g/mL x 1000 mL = 1230 g of solution

0.387 mol/L x 1 L = 0.387 mol HCl

0.387 mol HCl x (36.5 g / 1 mol) = 14.1 g HCl

mass of water = 1230 g solution - 14.1 g HCl = 1216 g H2O = 1.216 kg H2O

molality = mol HCl / kg water = 0.387 mol / 1.216 kg = 0.318 mol/kg (or 0.318 molal)

Answer 2

Answer:

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