##### Exercise | Q 2.07 | Page 45

# Answer the following in one or two sentences.

## Which of the following solution will have higher freezing point depression and why?

i. 0.1 m NaCl

ii. 0.05 m Al2(SO4)3

### Solution:

For 0.1 m NaCl:

NaCl → Na^{+} + Cl^{-}

0.1 m 0.1 m 0.1 m

Total particles in solution = 0.2 mol

For 0.05 m Al_{2}(SO_{4})_{3}:

Al_{2}(SO_{4})_{3} → 2Al^{3+} + 3$\text{SO}}_{4}^{2-$

0.05 m 0.1 m 0.15 m

Total particles in solution = 0.25 mol

Al_{2}(SO_{4})_{3} solution contains more number of particles than NaCl solution. Hence, Al_{2}(SO_{4})_{3} solution has maximum ΔT_{f}.

Therefore, the freezing point depression of 0.05 m Al_{2}(SO_{4})_{3} solution will be higher than 0.1 m NaCl solution.

For 0.1 m NaCl:

NaCl → Na^{+} + Cl^{-}

0.1 m 0.1 m 0.1 m

Total particles in solution = 0.2 mol

For 0.05 m Al_{2}(SO_{4})_{3}:

Al_{2}(SO_{4})_{3} → 2Al^{3+} + 3

0.05 m 0.1 m 0.15 m

Total particles in solution = 0.25 mol

Al_{2}(SO_{4})_{3} solution contains more number of particles than NaCl solution. Hence, Al_{2}(SO_{4})_{3} solution has maximum ΔT_{f}.

Therefore, the freezing point depression of 0.05 m Al_{2}(SO_{4})_{3} solution will be higher than 0.1 m NaCl solution.