What are the boiling point and freezing point of a 0.516 m aqueous solution of any nonvolatile, nonelectrolyte solute?

Boiling T° for solution is 100.2°C

Freezing T° for solution is  - 0.95°C

Explanation:

Formula for boiling point elevation:

Boiling T° for solution - Boiling T° for pure solvent = Kb . m

Formula for freezing point depression:

Freezing T° for pure solvent - Freezing T° for solution = Kf . m

Boiling T° for pure solvent = 100°C

Freezing T° for pure solvent = 0°C

Kb = Ebulloscopic constant for water → 0.512°C /m

Kf = Cryoscopic constant for water → 1.86°C / m

Let's replace data:

Boiling T° for solution = 0.512°C/m . 0.516 m + 100°C = 100.2°C

Freezing T° for solution = 1.86°C/m . 0.516 m - 0°C = - 0.95°C

if you're asking what comes out, it's steam.