The mathematical expression is given as:
[tex]\Delta T = k_{f}m[/tex] (1)
where, [tex]\Delta T[/tex] = depression in freezing point
[tex]k_{f}[/tex] = molal freezing point.
Now, first calculate the [tex]\Delta T = 25.50^{o}C -24.59^{o}C[/tex]
[tex]\Delta T= 0.91^{o}C[/tex]
Substitute the values in equation (1), we get
[tex]0.91^{o}C = 9.1^{o}C/m \times m[/tex]
[tex]m = \frac{0.91^{o}C}{9.1^{o}C/m}[/tex]
= [tex]0.1 molal[/tex] or [tex]\frac{0.1 moles of water}{kg of butanol}[/tex]
Now,
In 0.1 moles of water = 1 kg of butanol
So, 11.9 g of butanol = [tex]11.9 g butanol\times \frac{0.1 mol of water}{kg butanol}[/tex]
Convert gram into kilogram, (1 kg =1000 g)
= [tex]11.9 g butanol\times \frac{0.1 mol of water}{kg butanol}\times \frac{1 kg}{1000 g}[/tex]
= [tex]0.00119 mole[/tex]
Mass of water present in sample = [tex]0.00119 mole\times 18 g/mol[/tex]
= [tex]0.02142 g[/tex]
Hence, grams of water present in the sample = [tex]0.02142 g[/tex]
