Redox Reactions Question 3
Question 3 - 25 January - Shift 1
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 $10^{-2} mL$ (Nearest integer)
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Answer: (12)
Solution:
Formula: Mole-mole Analysis and Density
millimole of $NaOH=0.24 \times 25$
$\therefore \quad$ millimole of acid $=0.24 \times 25$
$\Rightarrow \quad$ mass of acid $=0.24 \times 25 \times 24.2 mg$
for pure acid,
$ V=\frac{W}{d} ;(d=1.21 kg / L^{n}=1.21 g / ml) $
$\therefore V=\frac{0.24 \times 25 \times 24.2}{1.12} \times 10^{-3}$
$ \begin{aligned} & =120 \times 10^{-3} ml \\ & =12 \times 10^{-2} ml \end{aligned} $
