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JEE Main 2021 (Online) 17th March Evening Shift
Numerical
+4
-1
JEE Main 2021 (Online) 17th March Evening Shift Chemistry - Some Basic Concepts of Chemistry Question 84 English
Consider the above reaction. The percentage yield of amide product is __________. (Round off to the Nearest Integer).

(Given : Atomic mass : C : 12.0 u, H : 1.0 u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u)
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2
JEE Main 2021 (Online) 17th March Morning Shift
Numerical
+4
-1
15 mL of aqueous solution of Fe2+ in acidic medium completely reacted with 20 mL of 0.03 M aqueous Cr2O$$_7^{2 - }$$. The molarity of the Fe2+ solution is __________ $$\times$$ 10-2 M. (Round off to the Nearest Integer).
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3
JEE Main 2021 (Online) 17th March Morning Shift
Numerical
+4
-1
JEE Main 2021 (Online) 17th March Morning Shift Chemistry - Some Basic Concepts of Chemistry Question 85 English
In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is ________%. (Round off to the Nearest Integer).

(Given atomic mass : C : 120 u, H : 1.0 u, O : 16.0 u, N : 14.0 u)
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4
JEE Main 2021 (Online) 17th March Morning Shift
Numerical
+4
-1
The mole fraction of a solute in a 100 molal aqueous solution is ___________ $$\times$$ 10$$-$$2. (Round off to the Nearest Integer).

[Given : Atomic masses : H : 1.0 u, O : 16.0 u ]
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