Question:
The molarity of the solution prepared by dissolving $6.3 \mathrm{~g}$ of oxalic acid $\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} .2 \mathrm{H}_{2} \mathrm{O}\right)$ in $250 \mathrm{~mL}$ of water in mol $\mathrm{L}^{-1}$ is $\mathrm{x} \times 10^{-2}$. The value of $\mathrm{x}$ is (Nearest integer)
[Atomic mass : $\mathrm{H}: 1.0, \mathrm{C}: 12.0, \mathrm{O}: 16.0]$
Solution:
$\left[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right]=\frac{\text { weight } / \mathrm{M}_{\mathrm{W}}}{\mathrm{V}(\mathrm{L})}$
$\Rightarrow x \times 10^{-2}=\frac{6.3 / 126}{250 / 1000}$
$x=20$
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