Question

The extent of ionization of NaCl and MgCl2 in water is 80% and 50% respectively. If a sample of water has 5.85% (w/w) NaCl and 9.5% (w/w) of MgCl2, then the mixture (Hint: Kb = 0.52Kkg/mol)

1)acts as an azeotrope
2)At constant pressure, the temperature change is less than 2K
3)At constant volume, the pressure change is more than 0.5%
4)The solubilty product decreases for the two salts.

Answer 1

Solution.

1.

Since the total mass fraction of salts is not equal to 100 %, we can say with almost complete confidence that this solution is not an azeotrope.

2.

$\Delta T = Kb \times Cm$

Find the number of moles of salts by mass fraction, based on the fact that the salt is contained in 84.65 g of water (100-5,85-9,50).

$n(NaCl) = \frac{5.85}{58.5} = 0.1 \ mol$

$n(MgCl2) = \frac{9.50}{95} = 0.1 \ mol$

Find the number of moles NaCl and MgCl2, based on the values of the degrees of dissociation of salts.

$\alpha = \frac{i-1}{k-1}$

$i = \alpha(k-1) + 1$

for NaCl, k = 2

for MgCl2, k = 3

i(NaCl) = 1.8

i(MgCl2) = 2

$n(NaCl) = 1.8 \times 0.1 = 0.18 \ mol$

$n(MgCl2) = 2 \times 0.1 = 0.2$

n(salts) = 0.2 + 0.18 = 0.38 mol

$Cm = \frac{n(salts) \times 1000}{m(H2O)}$

Cm = 4.49

$\Delta T =0.52 \times 4.49 = 2.33$

3.

$\Delta P = Po \times X \times i$

$\frac{\Delta P}{Po} = X$

$\frac{\Delta P}{Po} = 0.38$

4.

Since the pressure change does not occur sharply, the solubility of substances does not change.

Answer:

3)