What is the boiling point of a solution made by adding 6.69 g of magnesium chloride to 250.0 g of water? . use the formula of the salt to obtain i. (kb for water is 0.512 oc/m)

100.432ºC

Explanation:

Ebullioscopy is the elevation of the boiling point of a solvent that has a solute nonvolatile dissolved. The change in the temperature may be calculated by Raoult's law:

ΔT = Kb.W.i

Where ΔT is the temperature change, Kb is the ebullioscopy constant, W is the molality and i is the Van't Hoff factor, which determinates the particles that affect the proper.

The molality is:

W = m1/(M1xm2)

Where m1 is the solute mass (in g), M1 is the molar mass of the solute, and m2 is the mass of the solvent (in kg).

The Van't Hoff factor is the number of final particles divided by the number of initial particles. Magnesium is from froup 2, so it forms the cation Mg⁺², and chlorine ins from group 7 and forms the anion Cl⁻, the salt is MgCl₂ and dissociates:

MgCl₂  → Mg⁺²(aq) + 2Cl⁻(aq)

So, it has 1 particle in initial, and 3 in final (1  Mg⁺² and 2 Cl⁻). So:

i = 3/1 = 3.

The molar mass of MgCl₂ is: 24.3 + 2x35.5 = 95.3 g/mol, m1 = 6.69g, m2 = 0.250 kg, so:

W = 6.69/(95.3x0.250)

W = 0.281 m

Then:

ΔT = 0.512x0.281x3

ΔT = 0.432ºC

The normal boling point of water is 100ºC, so

T - 100 = 0.432

T = 100.432ºC

the answer is c. nacl