MCQ
The molarity of the solution prepared by dissolving $6.3\, \mathrm{~g}$ of oxalic acid $\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)$ in $250\, \mathrm{~mL}$ of water in $\operatorname{mol} \mathrm{L}^{-1}$ is $\mathrm{x} \times 10^{-2} .$ The value of $\mathrm{x}$ is ..... . (Nearest integer)

[Atomic mass : $\mathrm{H}: 1.0, \mathrm{C}: 12.0, \mathrm{O}: 16.0]$

  • A
    $0.20$
  • B
    $2$
  • C
    $200$
  • $20$

Answer

Correct option: D.
$20$
d
${\left[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right]=\frac{\text { weight } / \mathrm{M}_{\mathrm{w}}}{\mathrm{V}(\mathrm{L})}}$

$\Rightarrow \mathrm{x} \times 10^{-2}=\frac{6.3 / 126}{250 / 1000}$

$\mathrm{x}=20$

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