The $$NaNO_3$$ 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]

We need to find the mass of $$NaNO_3$$ required to make 50 mL of solution containing 70.0 mg of $$Na^+$$ per mL.

The total mass of $$Na^+$$ required in 50 mL is $$50 \times 70.0 = 3500$$ mg $$= 3.5$$ g.

The molar mass of Na is 23 g/mol, so the moles of $$Na^+$$ needed are $$\frac{3.5}{23} = \frac{35}{230} = \frac{7}{46}$$ mol.

Since each formula unit of $$NaNO_3$$ contains one $$Na^+$$ ion, the moles of $$NaNO_3$$ required equal the moles of $$Na^+$$, which is $$\frac{7}{46}$$ mol.

The molar mass of $$NaNO_3 = 23 + 14 + 3 \times 16 = 23 + 14 + 48 = 85$$ g/mol.

The mass of $$NaNO_3$$ required is $$\frac{7}{46} \times 85 = \frac{595}{46} = 12.93$$ g.

Rounded off to the nearest integer, the answer is $$13$$ g.