Question

0.25 g of an organic compound containing chlorine gave 0.40 g of silver chloride in Carius estimation. The percentage of chlorine present in the compound is ______. [in nearest integer]
(Given : Molar mass of \(Ag\) is 108 g mol^–1 and that of \(Cl\) is 35.5 g mol^–1

Answer 1

Correct Answer: 40

To determine the percentage of chlorine in the organic compound, we start with the reaction producing silver chloride (AgCl) from chlorine (Cl) in the compound:
Step 1: Calculate moles of AgCl:
\[ \text{Molar mass of AgCl} = \text{Molar mass of Ag} + \text{Molar mass of Cl} = 108 \, \text{g/mol} + 35.5 \, \text{g/mol} = 143.5 \, \text{g/mol} \]
\[ \text{Moles of AgCl} = \frac{0.40 \, \text{g}}{143.5 \, \text{g/mol}} \approx 0.0028 \, \text{mol} \]
Step 2: Since each mole of AgCl contains one mole of Cl, the moles of Cl are 0.0028 mol.
Step 3: Calculate mass of Cl:
\[ \text{Mass of Cl} = 0.0028 \, \text{mol} \times 35.5 \, \text{g/mol} = 0.0994 \, \text{g} \]
Step 4: Calculate percentage of chlorine in the compound:
\[ \text{Percentage of Cl} = \left( \frac{0.0994 \, \text{g}}{0.25 \, \text{g}} \right) \times 100\% = 39.76\% \]
Thus, the percentage of chlorine in the compound, rounded to the nearest integer, is 40. The calculated percentage is 39.76%, which confirms it falls within the given range of 40.