Question

A 50.0-mL solution is initially 1.49% MgCl2 by mass and has a density of 1.05 g/mL.

What is the freezing point of the solution after you add an additional 1.34 g MgCl2? (Use i = 2.5 for MgCl2.)

Express your answer using two significant figures.

Answer 1

Solution-

( 50.0 mL) (1.05 g/mL) = 52.5 grams of solution

(1.49% by mass) (52.5 grams of solution) = 0.7822 grams of MgCl2

52.5 grams of solution - 0.7822 grams of MgCl2 = 51.72 g H2O

0.7822 grams of MgCl2 + an additional 1.34 g MgCl2 = 2.1222 grams of MgCl2 total

Now we use molar mass and find find moles
(2.1222 grams of MgCl2) (1 mole MgCl2 / 95.21 grams) = 0.0222896 moles MgCl2

51.34 grams of solution - 0.7822 grams of MgCl2 = 50.56 g H2O

Now ,the molality is
( 0.0222896 moles MgCl2) / 0.05172 kg water = 0.4309 molal
Now,
dT = i Kf (molality)

dT = 2.5 (1.86 C / molal) ( 0.4309 molal)

dT = 2.00 C drop in temp

Hence the new freezing point will be
- 2.0 Celsius