Solve the following


The $\mathrm{NaNO}_{3}$ weighed out to make $50 \mathrm{~mL}$ of an aqueous solution containing $70.0 \mathrm{mgNa}^{+}$per $\mathrm{mL}$ is ___________ g.(Rounded off to the nearest integer)

[Given: Atomic weight in $\mathrm{g}$ mol $^{-1}$. Na: $23 ; \mathrm{N}: 14 ; \mathrm{O}: 16$ ]



$\mathrm{Na}^{+}=70 \mathrm{mg} / \mathrm{mL}$

$\mathrm{W}_{\text {Nat }}$ in $50 \mathrm{~mL}$ solution $=70 \times 50 \mathrm{mg}$

$=3500 \mathrm{mg}$

$=3.5 \mathrm{gm}$

Moles of $\mathrm{Na}^{+}$in $50 \mathrm{ml}$ solution $=\frac{3.5}{23}$

Moles of $\mathrm{NaNO}_{3}=$ moles of $\mathrm{Na}^{+}$

$=\frac{3.5}{23} \mathrm{~mol}$

Mass of $\mathrm{NaNO}_{3}=\frac{3.5}{23} \times 85=12.934$

$\simeq 13 \mathrm{gm}$ Ans.

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