Question

A

pure solvent freezes at 12.0 C. A solution of 0.980 g of the solute
and 13.870 g of solvent froze at 5.1 C. The molar mass of the
solute is 178.2 g/mol. Calculate the freezing point depression
constant, Kf for the solvent.

Answer 1

`K_(f)` for solvent is
`17^(0)\textrm{C}.kg.mol^(-1)`

Step-by-step explanation:

• Let's assume that the solute is non-volatile as well as non-electrolyte.
• For a solution with non-volatile solute and non-electrolyte solute-

`\Delta T_(f)=K_(f).m`, where
`\Delta T_(f)` is depression in freezing point and m is molality of solution

Molality of solution (m) = (moles of solute/mass of solvent in kg)

=
`((0.980)/(178.2))/(0.01387)mol/kg`

= 0.396 mol/kg

`\Delta T_(f)=(12.0-5.1)^(0)\textrm{C}=6.9^(0)\textrm{C}`

So,
`K_(f)=(\Delta T_(f))/(m)=(6.9)/(0.396)^(0)\textrm{C}.kg.mol^(-1)=17^(0)\textrm{C}.kg.mol^(-1)`