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\(0 .3780 \mathrm{~g}\) of an organic chloro compound gave \(0.5740\) \(g\) of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
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Here, the mass of the substance taken \(=0.3780 \mathrm{~g}\)
Mass of \(\mathrm{AgCl}\) formed \(=0.5740 \mathrm{~g}\)
Now 1 mole of \(\mathrm{AgCl} \equiv 1 \mathrm{~g}\) atom of \(\mathrm{Cl}\) or \((108+35.5)=143.5 \mathrm{~g}\) of \(\mathrm{AgCl} \equiv 35.5 \mathrm{~g}\) of \(\mathrm{Cl}\) Applying the relation, Percentage of chlorine \(=\frac{35.5}{143.5} \times \frac{\text { Mass of } \mathrm{AgCl} \text { formed }}{\text { Mass of substance taken }} \times 100\) \(=\frac{35.5}{143.5} \times \frac{0.5740}{0.3780} \times 100=37.56 \%\)
Mass of \(\mathrm{AgCl}\) formed \(=0.5740 \mathrm{~g}\)
Now 1 mole of \(\mathrm{AgCl} \equiv 1 \mathrm{~g}\) atom of \(\mathrm{Cl}\) or \((108+35.5)=143.5 \mathrm{~g}\) of \(\mathrm{AgCl} \equiv 35.5 \mathrm{~g}\) of \(\mathrm{Cl}\) Applying the relation, Percentage of chlorine \(=\frac{35.5}{143.5} \times \frac{\text { Mass of } \mathrm{AgCl} \text { formed }}{\text { Mass of substance taken }} \times 100\) \(=\frac{35.5}{143.5} \times \frac{0.5740}{0.3780} \times 100=37.56 \%\)
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