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Ways to Define Concentration

The NaNO3 wei...

Question

The $N a N O_{3}$ weighed to make $50 m L$ of an aqueous solution containing 70.0 $m g$ $N a^{+}$ per $m L$ is _____ $g$ . (Rounded off to the nearest integer)

[Given: Atomic weight in $g m o l^{- 1}$ . Na: 23; N: 14; O: 16.]

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Solution

$N a^{+}$ = 70 $m g m L^{- 1}$

$W_{N a}$ in 50 $m L$ solution = 70 × 50 $m g$ = 3500 $m g$ = 3.5 $g$

Moles of $N a^{+}$ in 50 $m L$ solution = $\frac{3.5}{23}$

Moles of $N a N O_{3}$ = moles of $N a^{+}$ = $\frac{3.5}{23} m o l$

Mass of $N a N O_{3} = \frac{3.5}{23} \times 85 = 12.94 g$
Mass of $N a N O_{3} = 13 g$

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