When sodium fusion extract of an organic compound was treated with CHCl$_3$, then violet colour of halogen appears. If 0.15 gm of the organic compound gives 0.12 gm of the silver halide, then find the percentage of halide in the organic compound.

Question

A first row transition metal (M) does not liberate $ \mathrm{H_2} $ gas from dilute HCl.

1 mol of aqueous solution of $ \mathrm{MSO_4} $ is treated with excess aqueous KCN and then $ \mathrm{H_2S(g)} $ is passed through the solution.

The amount of $ \mathrm{MS} $ (metal sulphide) formed from the above reaction is ________ mol.

Answer 1

Statement-I Analysis: Sucrose is dextrorotatory. Hydrolysis produces D-Glucose (dextro) and D-Fructose (laevo). The specific rotation of Fructose ($-92.4^\circ$) has a larger magnitude than that of Glucose ($+52.5^\circ$). The resulting mixture has a net negative rotation (laevorotatory). This phenomenon is called Inversion of Cane Sugar. Statement I is true. Statement-II Analysis: The statement claims "levorotation of glucose is more than dexrotation of fructose". This is factually incorrect and physically impossible (Glucose is dextro, Fructose is laevo). The correct logic is that the **laevorotation of fructose** is greater in magnitude than the **dextrorotation of glucose**. Thus, Statement II is false. Conclusion: Statement I is True, Statement II is False.

When sodium fusion extract of an organic compound was treated with CHCl$_3$, then violet colour of halogen appears. If 0.15 gm of the organic compound gives 0.12 gm of the silver halide, then find the percentage of halide in the organic compound.

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The Correct Option is A

Solution and Explanation

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