Question

$0.1914g$ of an organic acid is dissolved in about $20ml$ of water. $25ml$ of $0\cdot 12NNaOH$ is required for the complete neutralization of the acid solution. The equivalent weight of the acid is:

Answer 1

$Molesofacid=\frac{0.1914}{M}MolesofNaOH=25\times {10}^{-3}\times 0.12=3\times {10}^{-3}Now,moleofacid=moleofNaOH\frac{0.1914}{M}=3\times {10}^{-3}M=\frac{0.1914}{3\times {10}^{-3}}=63.8g{mol}^{-1}$\\