How are $0.50 mol N a_{2} C O_{3} and 0.50 M N a_{2} C O_{3}$ different?

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Solution

$M o l a r m a s s o f N a_{2} C O_{3} = (2 \times 23) + (1 \times 12) + (3 \times 16) = 106 g m o l^{- 1}$

Now,
$1 mole of N a_{2} C O_{3} means 106 g of N a_{2} C O_{3}$
$\Rightarrow 0.5 m o l e o f N a_{2} C O_{3} = 0.5 \times 106 = 53 g N a_{2} C O_{3}$
Further, $0.50 M of N a_{2} C O_{3} means 0.50 mol of N a_{2} C O_{3}$ in $1 L$ of water, which means $53 g of N a_{2} C O_{3} is present in 1 L of water$

$0.5 m o l$ is the amount but $0.50 M is 0.50 m o l / L$ i.e., it is not the amount but the concentration.

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