1. An aqueous solution was prepared by mixing 70 g of an unknown nondissociating solute into 100 g of water. The solution has a boiling point of 101.11°C. What is the molar mass of the solute?

The molar mass of the solute is 322.9 g/mol

Explanation:

Boiling point elevation to solve this:

ΔT = Kb . m

ΔT = T° boiling of solution - T°boiling of pure solvent

Kb → Ebullioscopic constant, for water is 0.512°C / m

m → molality

101.11°C - 100°C = 0.512 °C/m . m

1.11°C / 0.512 m/°C = m

2.17 = m → These are the moles of solute in 1kg of solvent

1kg = 1000 g. Let's make the rule of three to determine the moles in our solvent volume:

In 1000 g we have 2.17 moles of solute

In 100 g we may have (100 . 2.17)/1000 = 0.217 moles

The moles we obtained, are the moles for 70 g of mass.

Let's determine the molar mass: (g/mol)

70 g/ 0.217 mol = 322.9 g/mol