Question

An acidic solution was prepared by dissolving $1g$ of monobasic acid $HA$ in $100g$ of water. A depression of $0.145K$ was observed at the freezing point. $0.75g$ of this same acid requires $27mL$ of $\frac{\text{N}}{9}NaOH$ solution for complete neutralization. Calculate the degree percentage of dissociation of this acid.
[Given: ${\left(\text{K}\right)}_{\text{f}}\left({\left(\text{H}\right)}_2\text{O}\right)={\left(\text{1.86 K kg mol}\right)}^{-1}$ ]

• A. $34\%$
• B. $68\%$
• C. $20\%$
• D. $17\%$

Answer 1

Answer: (D)

Write the given parameters and calculate the required parameters using the following expression:
$\Delta {\text{T}}_{\text{f}}={\text{K}}_{\text{f}}=\frac{{\text{W}}_{\text{acid}}}{{\text{M}}_{\text{acid}}\times {\text{W}}_{{\text{H}}_2\text{O}}}\times 1000$
$\Delta {\text{T}}_{\text{f}}={\text{K}}_{\text{f}}\times \frac{{\text{W}}_{\text{acid}}}{{\text{M}}_{\text{acid}}\times {\text{W}}_{{\text{H}}_2\text{O}}}\times 1000$
$\text{0.145}=\text{1.86}\times \frac{\text{1.0}}{{\text{M}}_{\text{acid}}\times 100}\times 1000$
$\text{0.145}=\text{1.86}\times \frac{10}{{\text{M}}_{\text{acid}}}$
${\text{M}}_{\text{acid}}=\frac{\text{18.6}}{\text{0.145}}=\text{128.28}$
${\left(\text{M}\right)}_{\text{acid}}(observed)=\text{128.28}$
$$\begin{gathered} M_{acid(normal)}\{\frac{0.75}{M_{acid}}\times 1000=27\times \frac{1}{9} \\ {M}_{acid}(normal)=150\} \end{gathered}$$
$\text{i}=\frac{150}{\text{128.28}}=\text{1.169}\approx \text{1.17}$
$\text{i}=1+\alpha$
$1+\alpha =\text{1.17}$
$\alpha =\text{0.17, \% }\alpha =\text{17\%}$