The ${NaNO}_3$ weighed out to make $50 mL$ of an aqueous solution containing 70.0mg 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, Weight of ${Na}^{+}$ in 1 ml solution = 70 g

Volume of solution = 50 ml

Weight of ${Na}^{+}in50mlofsolution=7050=3.5g$

Moles of ${Na}^{+}=\frac{3.5}{23}$

Moles of ${NaNO}_3=molesof{Na}^{+}=\frac{3.5}{23}$

Weight of ${NaNO}_3=\frac{3.5}{23}\times 85=13g$

Therefore, weight of ${NaNO}_3$ is 13 g.