MCQ
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)
  • A
    $6$
  • $12$
  • C
    $3$
  • D
    $24$

Answer

Correct option: B.
$12$
b
millimole of $NaOH =0.24 \times 25$

$\therefore \quad \text { millimole of acid }=0.24 \times 25$

$\Rightarrow \quad \text { mass of acid }=0.24 \times 25 \times 24.2\,mg$

$\text { for pure acid, }$

$\qquad V =\frac{ w }{ d } ;( d =1.21\,kg / L =1.21\,g / ml )$

$\therefore V =\frac{0.24 \times 25 \times 24.2}{1.12} \times 10^{-3}$

$=120 \times 10^{-3}\,ml$

$=12 \times 10^{-2}\,ml$

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