How many moles are present in $5.3 g$ of anhydrous sodium carbonate?

0.03

moles

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0.04

moles

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0.05

moles

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0.01

moles

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Solution

The correct option is A $0.05$ moles
given that
$5.3 g N a_{2} C O_{3}$
atomic mass $= 2 \times 23 + 12 + 3 \times 16$
$= 106 g / m o l$
Moles $= \frac{g r a m w e i g h t}{a t o m i c m a s s}$
$= \frac{5.3}{106} = 0.05 m o l e s$

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