Question

The freezing point (tf) for t-butanol is 25.50°c and kf is 9.1°c/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 11.9-g sample of t-butanol is measured as 24.59°c, how many grams of water are present in the sample?

Answer 1

The mathematical expression is given as:

`\Delta T = k_(f)m` (1)

where,
`\Delta T` = depression in freezing point

`k_(f)` = molal freezing point.

Now, first calculate the
`\Delta T = 25.50^(o)C -24.59^(o)C`

`\Delta T= 0.91^(o)C`

Substitute the values in equation (1), we get

`0.91^(o)C = 9.1^(o)C/m *  m`

`m = (0.91^(o)C)/(9.1^(o)C/m)`

=
`0.1 molal` or
`(0.1 moles of water)/(kg of butanol)`

Now,

In 0.1 moles of water = 1 kg of butanol

So, 11.9 g of butanol =
`11.9 g butanol* (0.1 mol of water)/(kg butanol)`

Convert gram into kilogram, (1 kg =1000 g)

=
`11.9 g butanol* (0.1 mol of water)/(kg butanol)* (1 kg)/(1000 g)`

=
`0.00119 mole`

Mass of water present in sample =
`0.00119 mole* 18 g/mol`

=
`0.02142 g`

Hence, grams of water present in the sample =
`0.02142 g`