Question

The freezing point of t-butanol is $\mathbf{25}\mathbf{.}\mathbf{50}\mathbf{°}\mathbf{C}$ and K_f is 9.18C kg/mol. Usually t-butanol absorbs water on exposure to air. If the freezing point of a 10.0-g sample of t-butanol is $\mathbf{24}\mathbf{.}\mathbf{59}\mathbf{°}\mathbf{C}$, how many grams of water are present in the sample?

Short answer

The amount of water that will be present in the 10 g or 0.1 mole sample of t-butanol is 0.018g. It can be determined by the relation of depression in freezing point i.e.,_{$\Delta {\text{T=k}}_{\text{f}}\times \text{m}$}

Step-by-step solution

Step 1:

According to the question, it is given that:

_{$\begin{array}{l}\Delta {\text{T}}_1\text{=25.50}°\text{C}\\ \Delta {\text{T}}_2\text{=24.59}°\text{C}\\ {\text{K}}_f\text{=9.18}°\text{Ckg/mol}\end{array}$}

Depression of freezing point will be: _{$\Delta {\text{T}}_1\text{-}\Delta {\text{T}}_2\text{=25.50-24.59}°\text{C=0.91}°\text{C}$}

Mass of solute = 10g

The depression in freezing point is given by _{$\Delta {\text{T=k}}_{\text{f}}\times \text{m}$}

Thus,

_{$\begin{array}{l}\text{0.91=9.18}\times \text{m}\\ \text{m=}\frac{0.91}{\text{9.18}}\text{=0.1molal}\end{array}$}

Step 2:

Now, the 0.1 molal refers to that 1 kg of t-butanol is resent in 0.1 molal solution.

By unitary method, we can determine that 10g of t-butanol will have

$\frac{\text{0.01}\times \text{10}}{1000}{\text{=10}}^{-3}\text{mole}$

Therefore, the amount of water presents in ${\text{10}}^{-3}\text{mole}$of t-butanol will be

_{${\text{10}}^{-3}\times \text{18=0.018g}$}