A research scholar requires 50 mL aqueous ${NaNO}_{3}$ solution containing $70.0mg{Na}^{+}$per mL. T

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A research scholar requires 50 mL aqueous ${NaNO}_{3}$ solution containing $70.0mg{Na}^{+}$per mL. The amount of ${NaNO}_{3}$ required is

Accepted Answer

Sol. Key concept: Convert the required mass of an ion to the mass of its salt.The total sodium-ion mass required is 50 mL × 70.0 mg mL⁻¹ = 3500 mg = 3.50 g. Each mole of NaNO₃ supplies one mole of Na⁺.$$M(NaNO_{3}) = 23 + 14 + 3(16) = 85\ g\ mol^{- 1}$$$$m(NaNO_{3}) = 3.50 	imes \frac{85}{23} = 12.93\ g$$The salt mass exceeds the sodium mass because nitrate also contributes to it. Hence (d).

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