Question

What mass of salt (nacl) should you add to 1.42 l of water in an ice cream maker to make a solution that freezes at -11.2 ∘c ? Assume complete dissociation of the nacl and density of 1.00 g/ml for water?

Answer 1

The mathematical expression for depression in freezing point is given as:

`\Delta T_(f) =i K_(f)* m`

where,

`\Delta T_(f)` = depression in freezing point

`K_(f)` = molal depression constant (
`1.86^(o)C/m`)

m = molality

i = Van't Hoff factor

`\Delta T_(f) =T_(solvent)-T_(solution)`

=
`0.0^(o)C-(-11.2^(o)C)`

=
`11.2^(o)C`

Now, put the values in formula,

`11.2^(o)C =2* 1.86^(o)C/m* m` (as sodium chloride dissociate into two ions, i =2)

m =
`(11.2^(o)C)/(1.86^(o)C/m)`

=
`6.02 m`

Now, molality of the solution =
`(number of moles of solute)/(kg of the solvent)`

Number of moles =
`(given mass in g)/(molar mass)`

Molar mass of sodium chloride = 58.44 g/mol

molality of the solution =
`((m in g)/(58.44 g/mol))/(1.42 kg)` (1 L = 1kg)

6.02 m =
`((m in g)/(58.44 g/mol))/(1.42 kg)`

mass in g =
`6.02 mol/kg * 58.44 g/mol * 1.42 kg`

=
`499.56 g`

Thus, mass of sodium chloride is
`499.56 g`