Question

Arrange the following aqueous solutions in terms of freezing point depression with the least depression first: 0.45m CaCl2, 0.45m glucose or 0.45m NaCl? Then determine the freezing point depression for each. For water, Kf=1.86 C/m

Answer 1

Step-by-step explanation:

Relation between
`k_(f)`, molality and temperature is as follows.

T =
`K_(f) * m * i`

It is also known as depression between freezing point where, i is the Van't Hoff factor.

Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.

i for
`CaCl_(2)` = 3

i for glucose = 1

i for NaCl = 2

Depression in freezing point will have a negative sign. Therefore, d

depression in freezing point for the given species is as follows.

`T_{CaCl_(2)} = -1.86 * 0.45 * 3`

=
`-2.511^(o)C`

`T_(glucose) = 1.86 * 0.45 * 1`

=
`-0.837^(o)C`

`T_(NaCl) = -1.86 * 0.45 * 2`

=
`-1.674^(o)C`

Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.

Glucose < NaCl <
`CaCl_(2)`