Question

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

Answer 1

Answer:

• 1. Boiling point: 100.264ºC

• 2. Freezing point: - 0.960ºC

Explanation:

1. Boling point

The boiling point of a solvent will increase when a solute is added. The boiling point elevation is a colligative property.

When a nonvolatile, nonelectrolyte solute is added to a solvent, the increase of the boiling point may be calculated using the formula:

        $\Delta T_b=m\times K_b$

Where m is the molality and Kb is the the molal boiling point constant (for water, Kb = 0.512ºC/m ).

Substitute and compute:

               $\Delta T_b=0.516m\times 0.512\º C/m\\\\\Delta T_b=0.264\º C$

Hence, add the increase in the boiling point to the normal boiling point of water: 100.000ºC

            $T_b=100.264\º C$

2. Freezing point

The freezing point of a solve will decrease when a solute is added. The depression on the freezing point is another colligative property.

When a nonvolatile, nonelectrolyte solute is added to a solvent, the depression of the boiling point may be calculated using the formula:

        $\Delta T_f=m\times K_f$

Where m is the molality and Kf is the the molal freezing point constant (for water, Kf = 1.86 ºC/m ).

Substitute and compute:

        $\Delta T_f=0.516m\times 1.86\º C/m\\\\\Delta T_f=0.960\º C$

       

Subtract the decrease on the freezing point from the normal freezing point of water: 0.000ºC

        $T_f=-0.960\º C$