The $\mathrm{NaNO}_{3}$ weighed out to make 50 mL of an aqueous solution containing $70.0 \mathrm{mg} \mathrm{Na}^{+}$ per mL is ____ g. (Rounded off to the nearest integer) [Given; Atomic weight in $\mathrm{g} \mathrm{~mol}^{-1}$ - Na: 23; N: 14; O: 16]

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Some Basic Concepts of Chemistry

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]

Answer: 13

Total Na

^{+}

in 50 mL:

70.0 \times 50 = 3500

= 3.5

g. Moles of Na

^{+}

\frac{3.5}{23} \approx 0.152

mol. Since each

NaNO_{3}

gives one Na

^{+}

, moles of

NaNO_{3} = 0.152

mol. Molar mass of

NaNO_{3} = 23 + 14 + 48 = 85

g/mol. Mass

= 0.152 \times 85 \approx 12.92

\approx 13

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer