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50 ec of oxalic acid is oxidised by 25 cc of $0.20 \mathrm{~N} \mathrm{KMnO}$. The mass of oxalic acid present in 500 ce of the solution is
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The correct answer is:
$3.15 \mathrm{~g}$
Equivalent moles of oxalic acid = Equivalent moles $=N_{1} V_{1}=25 \times 0.2 \times 10^{-3}=5 \times 10^{-3}=0.005$
$\frac{\text { Weight of oxalic acid }}{\text { Eq. weight of oxalic acid }}=0.005$
Weight of oxalic acid $=0.05 \times 63=0.315 \mathrm{~g}$ $\therefore 0.315 \mathrm{~g}$ in 50 ce of oxalic acid.
In 500 ce solution, mass of oxalic acid is
$$
=0.315 \times \frac{500}{50}=3.15 \mathrm{~g}
$$
$\frac{\text { Weight of oxalic acid }}{\text { Eq. weight of oxalic acid }}=0.005$
Weight of oxalic acid $=0.05 \times 63=0.315 \mathrm{~g}$ $\therefore 0.315 \mathrm{~g}$ in 50 ce of oxalic acid.
In 500 ce solution, mass of oxalic acid is
$$
=0.315 \times \frac{500}{50}=3.15 \mathrm{~g}
$$
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