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Saturday, 31 October

JEE Main 2024 · ChemistryMultiple choiceSingle correctEasyRecall

JEE Main 8 April 2024, Shift 2, Chemistry Q71

Question 71 of 90 in this shift, Chemistry question 11 of 30, Section A.

In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent. Which is further treated with nitric acid and ammonium molybdate respectively. The yellow coloured precipitate obtained is :
  1. (1)Na3PO4.12MoO3\mathrm{Na_3PO_4.12MoO_3}
  2. (2)(NH4)3PO4.12MoO3\mathrm{(NH_4)_3PO_4.12MoO_3}Official answer
  3. (3)(NH4)3PO4.12(NH4)2MoO4\mathrm{(NH_4)_3PO_4.12(NH_4)_2MoO_4}
  4. (4)MoPO4. 21 NH4NO3\mathrm{MoPO_4.\,21\,NH_4NO_3}

Official answer

Option 2

NTA final key.

Chapter
Organic Chemistry: Some Basic Principles and Techniques
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.