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QJEEM25Chab74: JEE Main 2025 Chemistry - Class 11 JEEM25Chab74 Solution: In Carius method for estimation of halogens, 180 mg of an…

In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl . The percentage composition of chlorine in the compound is _________ %. (Given : molar mass in g mol−1\mathrm{g} \mathrm{~mol}^{-1} of Ag:108,Cl:35.5\mathrm{Ag}: 108, \mathrm{Cl}: 35.5 )

Step-by-Step Verified Solution

To find the percentage composition of chlorine in an organic compound using the Carius method, we follow these steps:

Calculate Millimoles of AgCl:

Since 143.5 mg of AgCl is produced, and the molar mass of AgCl is 143.5 g/mol, the millimoles of AgCl is:

Millimoles of AgCl=143.5 mg143.5 mg/mmol=1 mmol\text{Millimoles of AgCl} = \frac{143.5 \, \text{mg}}{143.5 \, \text{mg/mmol}} = 1 \, \text{mmol}

Determine Millimoles of Cl:

In AgCl, there is a 1:1 ratio of Ag to Cl, so the millimoles of Cl are equal to that of AgCl:

Millimoles of Cl=1 mmol\text{Millimoles of Cl} = 1 \, \text{mmol}

Calculate Mass of Cl:

Using the molar mass of Cl (35.5 g/mol), the mass of Cl is:

Mass of Cl=35.5×10−3 g=35.5 mg\text{Mass of Cl} = 35.5 \times 10^{-3} \, \text{g} = 35.5 \, \text{mg}

Compute Percentage by Mass of Cl:

The total mass of the organic compound is 180 mg, so the percentage of chlorine is:

% Cl by mass=(35.5 mg180 mg)×100=19.72%\% \text{ Cl by mass} = \left( \frac{35.5 \, \text{mg}}{180 \, \text{mg}} \right) \times 100 = 19.72\%

Round to the Nearest Integer:

Rounding 19.72% to the nearest integer gives 20%.

Thus, the percentage composition of chlorine in the compound is approximately 20%.

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