Asolution prepared by dissolving 3.27 g of a nonvolatile, covalent solute in 30.0 g of water boils at 100.74 ºc. what is the approximate molar mass of the solute? k bp h 2o = 0.512 ºc/m.

75.2g/mol

Explanation:

The normal boiling point of water is 100°C and the boiling point of this solution is 100.74°C.

The elevation in the boiling point (ΔTb) can be calculated using the following equation.

ΔTb = Kb × b

where,

Kb is the ebulloscopic constant (Kb(H₂O) = 0.512°C/m)

b is the molality of the solution

b = ΔTb/Kb

b = (100.74°C-100°C)/(0.512°C/m)

b = 1.45 m

The mass of the solvent is 30.0 g (30.0 × 10⁻³ kg). The moles of solute are:

30.0 × 10⁻³ kg Solvent × (1.45 mol Solute/1 kg Solvent) =0.0435 mol Solute

0.0435 moles of solute have a mass of 3.27 g. The molar mass of the solute is:

M = 3.27g/0.0435 mol = 75.2g/mol