Question

1g of a monobasic acid in 100 g water lowers the freezing point by ${0.168}^o$C. If 0.2 g of the same acid requires 15 ml of $\frac{N}{10}$ alkali solution for complete neutralization, the degree of dissociation of acid is nearly:

($K_f$ for $H_{2}O=1.86Kkgmo{l}^{-1})$

• A. $20$ %
• B. $30$ %
• C. $40$ %
• D. $50$ %

Answer 1

The correct option is A $20$ %

$\Delta T_f=\frac{K_f\times W_B\times 1000}{m_B^{{}^{\prime }}\times W_A}$

Substituting the values in above equation we get,

$\therefore m_B^{{}^{\prime }}\cong 110.7(m_B^{{}^{\prime }}=\text{abnormal molecular mass of acid})$

Again,

$eq.ofacid=\text{eq. of alkali used}$

$\frac{0.2}{E_B}=15\times {10}^{-3}\times \frac{1}{10}$

$\therefore E_B=m_B\cong 133(\because acidismonobasic))$

$i=\frac{m_B}{m_B^{{}^{\prime }}}=1+(2-1)\alpha$

$1+a=\frac{133}{110.7}$

$\alpha =0.201$

"Or" $\alpha \%\approx 20\%$