Question

Show the calculation of the freezing point of a solution made by dissolving 12.6 grams of the electrolyte NaNO3 in 200 grams of water. Kf for water is 1.86 and the FP of pure water is 0oC.

Answer 1

Answer: The freezing point of the solution is
`-2.8^0C`

Explanation:-

Depression in freezing point is given by:

`\Delta T_f=i* K_f* m`

`\Delta T_f=T_f^0-T_f=(0-T_f)^0C` = Depression in freezing point

i= vant hoff factor = 2 (for
`NaNO_3` which dissociates to give two ions )

`K_f` = freezing point constant =
`1.86^0C/m`

m= molality

`\Delta T_f=i* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}`

Weight of solvent (water)= 200 g = 0.2 kg

Molar mass of electrolyte
`NaNO_3` = 85 g/mol

Mass of electrolyte added = 12.6 g

`(0-T_f)^0C=2* 1.86* (12.6g)/(85 g/mol* 0.2kg)`

`(0-T_f)^0C=2.8`

`T_f=-2.8^0C`

The freezing point of the solution is
`-2.8^0C`