JEE MAIN - Chemistry (2021 - 26th February Evening Shift - No. 25)
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]
[Given : Atomic weight in g mol$$-$$1 - Na : 23; N : 14; O : 16]
Answer
13
Explanation
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 = $${{3.5} \over {23}}$$
Moles of NaNO3 = moles of Na+
= $${{3.5} \over {23}}$$ mol
Mass of NaNO3 = $${{3.5} \over {23}} \times 85 = 12.934$$
$$ \simeq $$ 13 gm
WNa+ in 50 mL solution
= 70 $$\times$$ 50 mg
= 3500 mg
= 3.5 gm
Moles of Na+ in 50 mL solution = $${{3.5} \over {23}}$$
Moles of NaNO3 = moles of Na+
= $${{3.5} \over {23}}$$ mol
Mass of NaNO3 = $${{3.5} \over {23}} \times 85 = 12.934$$
$$ \simeq $$ 13 gm
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