$=0.5$ moles No. of moles $=$ molarity $\times$ volume

Requird mass of $\mathrm{HNO}_{3}=0.5 \times 63$

$=31.5 \mathrm{gm}$

$70 \mathrm{gm}$ of $\mathrm{HNO}_{3}$ are present in $100 \mathrm{gm}$ of solution,

So $1\; gm$ will be present in $100 / 70 \mathrm{gm}$ of solution.

$31.5 \mathrm{gm}$ be present in $(100 / 70) \times 31.5 \mathrm{gm}$ of solution

$=45 \;\mathrm{gm}$