What are the normal freezing points and boiling points of the following solutions? (a) 22.1 g NaCl in 129 mL of water.

The freezing point of solution is -10.9°C and the boiling point of solution is 103.°C

Explanation:

To calculate the mass of water, we use the equation:

Density of water = 1 g/mL

Volume of water = 129 mL

Putting values in above equation, we get:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

To calculate the depression in freezing point, we use the equation:

Or,

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2 (For NaCl)

= molal freezing point elevation constant = 1.86°C/m

= Given mass of solute (NaCl) = 22.1 g

= Molar mass of solute (NaCl) = 58.5  g/mol

= Mass of solvent (water) = 129 g

Putting values in above equation, we get:

Elevation in boiling point is defined as the difference in the boiling point of solution and boiling point of pure solution.

To calculate the elevation in boiling point, we use the equation:

Or,

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For NaCl)

= molal boiling point elevation constant = 0.512°C/m

= Given mass of solute (NaCl) = 22.1 g

= Molar mass of solute (NaCl) = 58.5  g/mol

= Mass of solvent (water) = 129 g

Putting values in above equation, we get:

Hence, the freezing point of solution is -10.9°C and the boiling point of solution is 103.°C

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