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?

The mathematical expression is given as:

            (1)

where, = depression in freezing point

= molal freezing point.

Now, first calculate the

Substitute the values in equation (1), we get

= or  

Now,

In 0.1 moles of water = 1 kg of butanol

So, 11.9 g of butanol =

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

=

=

Mass of water present in sample =

=

Hence, grams of water present in the sample  =