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  4. The amount of oxalic acid (mol. wt. 63 ) required to prepare 500 mL of its 0.10 N solution is

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Q. The amount of oxalic acid (mol. wt. $63$ ) required to prepare $500\, mL$ of its $0.10\, N$ solution is

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Solution:

$1000\, mL$ of $1\, N$ oxalic acid sol $=63\,g$

$500\, mL$ of $0.1 \,N$ oxalic acid sol

$=\frac{63}{1000} \times 500 \times 0.1=3.15 \,g$