1. Tardigrade
  2. Question
  3. Chemistry
  4. One given of a monobasic acid dissolved in 200 g of water lowers the freezing point by 0.186° C. On the other hand, when 1 g of the same acid is dissolved in water so as to make the solution 200 mL, this solution requires 125 mL of 0.1 N NaOH for complete neutralization. Calculate α for the acid in percent. ( K f =1.86 K kg mol -1)

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Q. One given of a monobasic acid dissolved in $200 \,g$ of water lowers the freezing point by $0.186^{\circ} C$. On the other hand, when $1 \,g$ of the same acid is dissolved in water so as to make the solution $200\, mL$, this solution requires $125 \,mL$ of $0.1 \,N\, NaOH$ for complete neutralization. Calculate $\alpha$ for the acid in percent. $\left( K _{ f }=1.86 \,K \,kg\, mol ^{-1}\right)$

Solutions

Solution:

$\frac{1}{M_{\text {acid }}}=\frac{125}{1000} \times 0.1$
$\Rightarrow M _{\text {acid }}=80\, g\, mol ^{-1}$
$0.186=(1+\alpha) \times 1.86 \times\left[\frac{\frac{1}{80}}{0.2}\right]$
$\Rightarrow \alpha=0.6 \Rightarrow 60 \%$