Question

A solution of water (kf=1.86 ∘c/m) and glucose freezes at − 4.15 ∘c. what is the molal concentration of glucose in this solution? assume that the freezing point of pure water is 0.00 ∘c.

Answer 1

The formula for depression of freezing point is:

`\Delta T_f = k_fm` -(1)

where,
`\Delta T_f` is depression of freezing point,
`k_f` is freezing point depression constant and
`m` is molality.

Freezing point of pure water =
`T_(o)^(f) = 0.00^(o) C` (given)

Freezing point of glucose =
`T_(f) = -4.15^(o) C` (given)

Depression of freezing point,
`\Delta T_f` =
`T_(o)^(f) - T_f`

Substituting the values:

`\Delta T_f = 0.00^(o) C - (-4.15^(o) C) = 4.15^(o) C`

Substituting the values in formula (1):

`4.15^(o) C = 1.86^(o)C/m* m`

`m = (4.15^(o) C)/(1.86^(o)C/m) = 2.231 m`

Hence, the the molal concentration of glucose in the given solution is
`2.231 m`.