Let's solve the problem step-by-step to find the percentage of element $$\mathrm{X}$$ and the formula of the violet-colored compound formed when the Lassaigne's extract of the organic compound is treated with sodium nitroprusside.

First, we'll calculate the percentage of the element $$\mathrm{X}$$ in the organic compound based on the mass of barium sulphate ($$\mathrm{BaSO_4}$$) precipitated.

The molecular weight of barium sulphate, $$\mathrm{BaSO_4}$$, is:

$$ \text{Molecular weight of } \mathrm{BaSO_4} = 137 + 32 + 64 = 233 \mathrm{~g/mol} $$

The atomic weight of sulphur (S) is 32 g/mol. When barium sulphate is formed, one mole of $$\mathrm{BaSO_4}$$ contains one mole of sulphur:

Moles of $$\mathrm{BaSO_4}$$ formed = Mass of $$\mathrm{BaSO_4}$$ / Molecular weight of $$\mathrm{BaSO_4}$$:

$$ \text{Moles of } \mathrm{BaSO_4} = \frac{1.2 \mathrm{~g}}{233 \mathrm{~g/mol}} = \frac{1.2}{233} \approx 0.00515 \mathrm{~mol} $$

Moles of sulphur = Moles of $$\mathrm{BaSO_4} = 0.00515 \mathrm{~mol}$$

The mass of sulphur in the compound is then:

$$ \text{Mass of sulphur} = \text{Moles of sulphur} \times \text{Atomic weight of sulphur} = 0.00515 \mathrm{~mol} \times 32 \mathrm{~g/mol} = 0.1648 \mathrm{~g} $$

Now, the percentage of element $$\mathrm{X}$$ (sulphur in this case) in the organic compound is:

$$ \text{Percentage of } \mathrm{X} = \left( \frac{\text{Mass of sulphur}}{\text{Mass of organic compound}} \right) \times 100 = \left( \frac{0.1648 \mathrm{~g}}{0.8 \mathrm{~g}} \right) \times 100 \approx 20.6\% $$

So, the percentage of element $$\mathrm{X}$$ is $$20.6\%$$.

Next, we need to determine the formula of the violet-colored compound formed when Lassaigne's extract is treated with sodium nitroprusside. The violet coloration is characteristic of the reaction between sodium nitroprusside and sulphur in the extract.

The chemical formula for the violet-colored compound formed in this reaction is:

$$ \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right] $$

Thus, the correct answer is:

Option B

$$ 20.6 \quad \& \quad \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right] $$

Physical Chemistry

Inorganic Chemistry

Organic Chemistry

Practical Organic Chemistry · Chemistry · COMEDK

MCQ (Single Correct Answer)