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

$0.91^{o}C = 9.1^{o}C/m \times m$  

$m = \frac{0.91^{o}C}{9.1^{o}C/m}$

= $0.1 molal$ or  $\frac{0.1 moles of water}{kg of butanol}$

Now,

In 0.1 moles of water = 1 kg of butanol

So, 11.9 g of butanol = $11.9 g butanol\times \frac{0.1 mol of water}{kg butanol}$

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

= $11.9 g butanol\times \frac{0.1 mol of water}{kg butanol}\times \frac{1 kg}{1000 g}$

= $0.00119 mole$

Mass of water present in sample = $0.00119 mole\times 18 g/mol$

= $0.02142 g$

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