JEE Main 17 March 2021 Shift 2 Solved Paper

Section: Chemistry

Question : 60 of 90

Marks: +1, -0

Consider the above reaction. The percentage yield of amide product is (Round off to the nearest integer). (Given : Atomic mass : C:12.0u,H:1.0u, N:14.0u,O:16.0u,Cl:35.5u )

[17 Mar 2021 Shift 2]

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Given mass=0.140g	0.388g	0.210g
Molar mass =140.5g/mol	169g/mol	273g/mol
No. of moles =0.001mol	0.0023mol	0.00077mol