JEE Main 2019ChemistryPractical ChemistryEasyMCQ

JEE Main 2019 — Practical Chemistry Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

An organic compound ‘A’ is oxidized with Na2O2 followed by boiling with HNO3. The resultant solution is then treated with ammonium molybdate to yield a yellow percipitate. Based on a above observation, the element present in the given compound is:

Choose an option

Show full solutionCorrect option: B
Correct answer
BPhosphorus

Step-by-step explanation

The phosphorus containing organic compounds are detected by 'Lassaigne's test' by heated with an oxidizing agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate.

The solution is boiled with nitric acid and then treated with ammonium molybdate to produced canary yellow precipitate. 

Phosphorus gets oxidised when it is treated with fuming conc. HNO3.
Na3PO4+3HNO3→H3PO4+3NaNO3

Phosphoric acid is used along with molybdenum compound for the identification of Magnesium.

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About this question

This is a previous-year question from JEE Main 2019, covering the Practical Chemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.