JEE Main 2021 — Some Basic Concepts Of Chemistry Question with Solution
From: JEE Main 2021 (Online) 26th February Evening Shift
Question
The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na+ per mL is _________ g. (Rounded off to the nearest integer)
[Given : Atomic weight in g mol1 - Na : 23; N : 14; O : 16]
[Given : Atomic weight in g mol1 - Na : 23; N : 14; O : 16]
Enter your answer
Show full solutionCorrect answer: 12.934
Correct answer
12.934
Step-by-step explanation
Na+ = 70 mg/mL
WNa+ in 50 mL solution
= 70 50 mg
= 3500 mg
= 3.5 gm
Moles of Na+ in 50 mL solution =
Moles of NaNO3 = moles of Na+
= mol
Mass of NaNO3 =
13 gm
WNa+ in 50 mL solution
= 70 50 mg
= 3500 mg
= 3.5 gm
Moles of Na+ in 50 mL solution =
Moles of NaNO3 = moles of Na+
= mol
Mass of NaNO3 =
13 gm
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This is a previous-year question from JEE Main 2021, covering the Some Basic Concepts Of 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.