JEE Main 2021ChemistrySome Basic Concepts of ChemistryMediumNumerical

JEE Main 2021 — Some Basic Concepts of Chemistry Question with Solution

JEE Main 2021 (26 Feb Shift 2)

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 gmol-1-Na : 23; N : 14; O : 16]

Enter your answer

Show full solutionCorrect answer: 13
Correct answer
13

Step-by-step explanation

Na+ present in 50 ml

=70 mg1 ml×50 ml=3500 mg=3.5 gm

moles of Na+=3.523= moles of NaNO3

weight of NaNO3=3.523×85=12.993 gm =13 gm

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Some Basic Concepts of Chemistry chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

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.