1. Find total mass of $Na^+$ required

Given concentration of sodium ions is $70.0\ \text{mg mL}^{-1}$ and total volume is $50\ \text{mL}$.

So total mass of $Na^+$ needed is:

$70.0\times 50 = 3500\ \text{mg} = 3.5\ \text{g}$

2. Relate $NaNO_3$ to $Na^+$

Each mole of $NaNO_3$ gives one mole of $Na^+$.

Molar mass of $NaNO_3$:

$23 + 14 + 3\times 16 = 23+14+48 = 85\ \text{g mol}^{-1}$

Thus, $85\ \text{g}$ of $NaNO_3$ contains $23\ \text{g}$ of $Na^+$.

3. Use mass ratio

If $23\ \text{g}$ of $Na^+$ comes from $85\ \text{g}$ of $NaNO_3$, then $3.5\ \text{g}$ of $Na^+$ will come from:

$\text{mass of } NaNO_3 = 3.5\times \frac{85}{23}$

$= \frac{297.5}{23} \approx 12.93\ \text{g}$

4. Round to nearest integer

$12.93 \approx 13$

So, the required mass of $NaNO_3$ is:

$\boxed{13\ \text{g}}$