A `HNO_(3)` solution (specific gravity=1.46) contains75% `HNO_(3)`. How much of this acid will be needed to dissolve 5g of CuO?

Explanation

Mass of 100 mL `HNO_(3)` solution`=100xx1.46=146g`
`underset((63.5+16)g=79.5g)(CuO(s))+underset(2xx63g=123g)(2HNO_(3)(g)) to Cu(NO_(3))_(2)(aq)+H_(2)O(l)`
79.5g CuO`-=126g" "HNO_(3)`
`therefore5g` `CuO-=7.92g" "HNO_(3)`
75g `HNO_(3)` is present in 146g solution.
`therefore 7.92g" "HNO_(3)` is present in `(146xx7.92)/(75)=15.417g`
solution.
`therefore 15.417g` acid will be required to dissolve 5 g of cupric oxide.