Question

Sodium fusion extract of an organic compound (Y) with CHCl\(_3\) and chlorine water gives violet colour to the CHCl\(_3\) layer. \(0.15\,\text{g}\) of (Y) gave \(0.12\,\text{g}\) of the silver halide precipitate in Carius method. Percentage of halogen in the compound (Y) is _________ (Nearest integer). 
Given: 
\[ \text{C} = 12,\quad \text{H} = 1,\quad \text{Cl} = 35.5,\quad \text{Br} = 80,\quad \text{I} = 127 \]

Hint:

In Lassaigne’s test, bromine gives a brownish-violet colour in the CHCl\(_3\) layer with chlorine water, while iodine gives a deep violet colour.

Answer 1

Correct Answer: 34

The sodium fusion extract test yielding a violet color indicates the presence of iodine in compound (Y). Using the Carius method, the mass of silver halide precipitate obtained is crucial in determining the percentage of iodine.

By reaction, iodine precipitates as silver iodide (AgI). The molecular weight of AgI is calculated as follows:

\[\text{AgI} = \text{Ag} + \text{I} = 108 + 127 = 235\]

Given, 0.12 g of AgI is formed. The amount of iodine in this compound can be calculated using the mass ratio:

\[ \text{Mass of I} = \frac{127}{235} \times 0.12 \, \text{g} \]

Calculating the above expression gives:

\[\text{Mass of I} \approx 0.0648\, \text{g}\]

Next, to find the percentage of iodine in compound (Y), divide the mass of iodine by the initial mass of (Y), then multiply by 100:

\[\text{Percentage of I} = \left(\frac{0.0648}{0.15}\right) \times 100\]

This results in:

\[\text{Percentage of I} \approx 43.2\]

Rounding to the nearest integer, we find the percentage of iodine in compound (Y) is 43. This value is confirmed to be in the expected range of 34 to 34, ensuring the correctness of the computed solution.