Verified Solution • 2025–26 Session moderate

QJEEM21Chbbb8: JEE Main 2021 Chemistry - Class 11 JEEM21Chbbb8 Solution: The NaNO 3 weighed out to make 50 mL of an aqueous solution…

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 mol −- 1 - Na : 23; N : 14; O : 16]

Step-by-Step Verified Solution

Na+ = 70 mg/mL

WNa+ in 50 mL solution

= 70 ×\times 50 mg

= 3500 mg

= 3.5 gm

Moles of Na+ in 50 mL solution = 3.523{{3.5} \over {23}}

Moles of NaNO3 = moles of Na+

= 3.523{{3.5} \over {23}} mol

Mass of NaNO3 = 3.523×85=12.934{{3.5} \over {23}} \times 85 = 12.934

≃\simeq 13 gm

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