How many grams of sodium oxalate are to be weighed to get $0.1$ mole of sodium oxalate $({N a}_{2} C_{2} O_{4})$ ?

13.4 g

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1.34 g

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134 g

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0.134 g

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Solution

The correct option is A $13.4 g$
given that $0.1$ moles of $N a_{2} C_{2} O_{4}$
Molecular weight $= 2 \times 23 + 2 \times 12 + 4 \times 16$
$= 134 g / m o l$
Moles $= \frac{g r a m w e i g h t}{M o l e c u l a r w e i g h t}$
$0.1$ moles $= \frac{g r a m w e i g h t}{134}$
gram weight $= 13.4 g$

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