Answer:
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
