When 800mL of 0 . 5M nitric acid is heated in a beaker, its volume...

The number of moles of nitric acid, ${n}_{{HNO}_{3}} = 0 . 5 \times 0 . 8$

$= 0 . 4mole$

${\left({n}_{{HNO}_{3}}\right)}_{remains} = 0 . 4 - \dfrac{11 . 5}{63}$

$= 0 . 4 - 0 . 1825$

$= 0 . 2175$

Molarity $= \dfrac{0 . 2175}{400} \times 1000$

$= \dfrac{0 . 2175}{0 . 4}$

$= 0 . 5437mole / lit .$

$≃ 0 . 54mole / lit .$

$= 54 \times {10}^{- 2}mol / lit .$