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:
$\Delta \text{T}_{\text{f}}=\text{K}_{\text{f}}=\frac{\text{W}_{\text{acid}}}{\text{M}_{\text{acid}} \times \text{W}_{\text{H}_{2} \text{O}}}\times 1000$
$\Delta \text{T}_{\text{f}}=\text{K}_{\text{f}}\times \frac{\text{W}_{\text{acid}}}{\text{M}_{\text{acid}} \times \text{W}_{\text{H}_{2} \text{O}}}\times 1000$
$\text{0.145}=\text{1.86}\times \frac{\text{1.0}}{\text{M}_{\text{acid}} \times 100}\times 1000$
$\text{0.145}=\text{1.86}\times \frac{10}{\text{M}_{\text{acid}}}$
$\text{M}_{\text{acid}}=\frac{\text{18.6}}{\text{0.145}}=\text{128.28}$
$\left(\text{M}\right)_{\text{acid}}\left(\right.observed\left.\right)=\text{128.28}$
$M_{acid \left(\right. normal \left.\right)}\left\{\right. \frac{0 .75}{M_{acid}}\times 1000=27\times \frac{1}{9} \\ M_{acid}\left(\right. normal \left.\right)=150 \left.\right\}$
$\text{i}=\frac{150}{\text{128.28}}=\text{1.169}\approx\text{1.17}$
$\text{i}=1+\alpha $
$1+\alpha =\text{1.17}$
$\alpha =\text{0.17, \% }\alpha =\text{17\%}$