Question

The density of a monobasic strong acid (Molar mass $24.2 \,g$ mol) is $1.21 \,kg \,L$. The volume of its solution required for the complete neutralization of $25 \,mL$ of $0.24\, M\, NaOH$ is ___$\times 10^{-2} mL$ (Nearest integer)

Answer 1

millimole of $NaOH =0.24 \times 25$
$\therefore $ millimole of acid $=0.24 \times 25$
$ \Rightarrow $ mass of acid $=0.24 \times 25 \times 24.2\, mg$
for pure acid,
$V =\frac{ w }{ d } ;( d =1.21 kg / L =1.21 g / ml )$
$\therefore V = \frac{0.24 \times 25 \times 24.2}{1.12} \times 10^{-3} $
$ =120 \times 10^{-3} ml $
$ =12 \times 10^{-2} ml$