How much $N a N O_{3}$ must be weighed out to make $100 m L$ of an aqueous solution containing $23 m g o f N a^{+} p e r m L$ ?

123.94 g

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80.5 g

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10.934 g

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8.5 g

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Solution

The correct option is D $8.5 g$
Molecular weight of $N a N O_{3} = 85$
$23 m g o f N a^{+} a r e p r e s e n t i n 1 m L$
$100 m L$ of solution contains $100 \times 23 = 2300 m g$
= $2.3 g N a^{+} i o n$
$23 g$ of $N a^{+}$ are present in $85 g$ of $N a N O_{3}$
$2.3 g$ of $N a^{+}$ are present in $\frac{85}{23} \times 2.3$
$= 8.5 g$ of $N a N O_{3}$

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