No of moles `HNO_(3)=("Mass of "HNO_(3))/("Molar mass of HNO"_(3))=(W)/(("63 g mol"^(-1)))`
Volume of solution in litres = 0.250 L
Molarity of solution `= ("No. of moles of HNO"_(3))/("Molarity of solution in litres")`
`(2.0 "mol L"^(-1))=(W)/(("63 g mol"^(-1))xx("0.250 L"))`
`W=("2.0 mol L"^(-1))xx("63 g mol"^(-1))xx(0.250 L)=31.5 g`
Actual mass of `HNO_(3)=(31.5g)xx(100)/(70)=45g`.