Question:
In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The percentage of chlorine in the organic compound is
In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The percentage of chlorine in the organic compound is
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Formula for Carius estimation: $\%X = \frac{At. Mass}{Mol. Mass(AgX)} \times \frac{W_{AgX}}{W_{org}} \times 100$.
Updated On: Feb 5, 2026
- 37.57\%
- 34.79\%
- 53.58\%
- 87.65\%
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The Correct Option is C
Solution and Explanation
Percentage of Chlorine $= \frac{\text{Atomic Mass of Cl}}{\text{Molar Mass of AgCl}} \times \frac{\text{Mass of AgCl}}{\text{Mass of Compound}} \times 100$.
Molar Mass of AgCl $= 108 + 35.5 = 143.5 \text{ g/mol}$.
$\% Cl = \frac{35.5}{143.5} \times \frac{0.5253}{0.2425} \times 100$.
$\% Cl = 0.24738 \times 2.1662 \times 100 \approx 53.58\%$.
Molar Mass of AgCl $= 108 + 35.5 = 143.5 \text{ g/mol}$.
$\% Cl = \frac{35.5}{143.5} \times \frac{0.5253}{0.2425} \times 100$.
$\% Cl = 0.24738 \times 2.1662 \times 100 \approx 53.58\%$.
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