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A sample containing 0.4775g of $(NH_4)_2C_2O_4$ and inert materials was dissolved in water and made strongly alkaline with KOH ,which converted $NH_4^+$ to $NH_3$.The liberated ammonia was distilled into exactly 50.0ml of 0.05035M $H_2SO_4$.The excess $H_2SO_4$ was back titrated with 11.3ml of 0.1214M $NaOH$.Calculate percent of N,percent of $(NH_4)_2C_2O_4$.(Mol.wt of $(NH_4)_2C_2O_4=124.10$ and at .wt.of N=14.0078)

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

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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.
answered Jan 7, 2014 by sreemathi.v
 

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