Question

an aqueous solution contains 5% by mass of urea and 10% by mass of sucrose.if molal depression of water is 1.86 K kg mol‐¹ then freezing point of solution is
a.-1.43⁰C
b.-2.43⁰C
b.-3.43⁰C
c.-4.43⁰C​

Answer 1

Answer:

Mass of solution = 100 g

Mass of glucose = 10 g

Mass of urea = 5 g

Therefore, mass of water = (100-10-5)=85 g

Total number of moles of solute =

5/60+10/342=0.112 mol

Effective molality of solution = 0.112 into 1000 / 85 = 1.31 m

1.86 into 1.31 = 2.43

Hence , the freezing point is 2.43 degree C