What is the molality of a solution of 1.25 moles of sugar dissolved in 0.750 kg of water? What is the boiling point and freezing point of the resulting solutions?

1.67 m is the molality of a solution .

The boiling point of a solution is 100.85°C.

The freezing point of a solution is -3.1°C.

Explanation:

Moles of sugar = 1.25 mol

Mass of water = 0.750 kg

Molality of the solution ;

1.67 m is the molality of a solution .

Freezing point of the resulting solution

where,

=elevation in boiling point =

T' = Boiling point of pure solvent

= boiling point of solution

= boiling point constant

m = molality

we have :

of water = 0.512 °C/m ,

m = 1.67 m

Boiling point of pure water = T' =  100°C

Boiling  point of solution =

The boiling point of a solution is 100.85°C.

Freezing point of the resulting solution

where,

=depression in freezing point =

T = Freezing point of pure solvent

= Freezing point of solution

= freezing point constant

m = molality

we have :

of water = 1.86°C/m ,

m = 1.67 m

Freezing point of pure water = T =  0°C

Freezing point of solution =

The freezing point of a solution is -3.1°C.

c. 4.54 x 10–12 j happens to be correct