Answer to Question #68277 in General Chemistry for kaitlyn

Question

a 5.00 M aqueous soln of magnesium chloride is made. the density of the soln is 1.07 g/ml. what is the boiling point of the soln? (kb=0.51 degrees celcius/m

Answer 1

Solution:

According to Raul’s law, the boiling point of the electrolyte solution is:

Tb(sln) – Tb(solv) = I· Kb·m (1),

Where I – isotonic coefficient; Kb – ebullioscopic constant; m – molality.

If we know molar concentration of the solution, so we can calculate molality of the same

solution, using formula:

С = 1000*ρ*m / (mM+1000), where C is molarity, and m is molality; ρ is density; M is molar

mass (M(MgCl 2 ) = 24 + 2·35.5 = 95 g/mol).

Thus, molality equals:

5.00 = 1000·1.07·m / (m·95 +1000)

5.00 = 1070·m / (m·95 +1000)

1070·m = 475·m + 5000

595·m = 5000

m = 8.4 mol/kg

The dissociation process of MgCl 2 :

MgCl 2 → Mg +2 + 2Cl - (number of produced ions is 3 (n = 3))

We assume that dissociation is 100%, so α = 1

Thus, we can calculate the number i:

I = 1 + α(n – 1) = 1 + 1(3-1)= 3

Using equation (1), we can define Tb(sln):

Tb(sln) = Tb(solv) + I· Kb·m

Tb(sln) = 100 + 3· 0.51·8.4 = 112.85 ( o C)

Answer: Tb(sln) = 112.85 o C.