Question

If you had 0.956 moles of a nonelectrolyte solute in 345 grams of water. What is the new freezing point? (The Kf of water is 1.86 °C/m )

Answer 1

Formula used in calculating freezing point is

$\Delta Tf=iKfm$

Molarity of the solution

$M=\dfrac{m}{v}$

no.of moles m = 0.956

volume of water = 0.345 kg

$M=\dfrac{0.956}{0.345}$ =2.77

Kf = 1.86

Therefore;

$\Delta Tf=1.86*2.77$ =5.15

The freezing point is 5.15⁰ C