Question:
The $\mathrm{NaNO}_{3}$ weighed out to make $50 \mathrm{~mL}$ of an aqueous solution containing $70.0 \mathrm{mg} \mathrm{Na}^{+}$per $\mathrm{mL}$ is_________ g. (Rounded off to the nearest integer)
[Given : Atomic weight in $\mathrm{g} \mathrm{mol}^{-1}-\mathrm{Na}: 23$;
$\mathrm{N}: 14 ; \mathrm{O}: 16]$
Solution:
$\mathrm{Na}^{+}$present in $50 \mathrm{ml}$
$=\frac{70 \mathrm{mg}}{1 \mathrm{ml}} \times 50 \mathrm{ml}=3500 \mathrm{mg}=3.5 \mathrm{gm}$
moles of $\mathrm{Na}^{+}=\frac{3.5}{23}=$ moles of $\mathrm{NaNO}_{3}$
weight of $\mathrm{NaNO}_{3}=\frac{3.5}{23} \times 85=12.993 \mathrm{gm}$
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