$(a)\;11.2\%,48.5\%\qquad(b)\;11.4\%,49.5\%\qquad(c)\;9.3\%,46\%\qquad(d)\;10.74\%,47.8\%$

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Meq of $(NH_4)_2C_2O_4$=Meq of $NH_4$=Meq of $H_2SO_4$

Used $\large\frac{w}{\Large\frac{124.10}{2}}$$\times 1000$=Meq of $NH_3$

$\Rightarrow (0.05035\times 2\times 50-11.3\times 0.1214)$

$\Rightarrow 3.663$

$w_{(NH_4)_2C_2O_4}=0.2273$

$\therefore \%$ of $(NH_4)_2C_2O_4=\large\frac{0.2273}{0.4775}$$\times 100=47.8\%$

Also Meq of N=Meq of $NH_3$

$\large\frac{w}{14}$$\times 1000=3.663$

$w_{N_2}=0.0513$

% of N=$\large\frac{0.0513}{0.4775}$$\times 100$

$\Rightarrow 10.74\%$

Hence (d) is the correct answer.

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