When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is $x \times 10^{- 2} M$ . (Nearest Integer)
(Molar mass of nitric acid is 63 g $m o l^{- 1}$ )

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answer is 55.

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Detailed Solution

Moles of nitric acid $= 800 \times 10^{- 3} \times 0.5 = 0.4$
Moles of nitric acid evaporated $= \frac{11.5}{63} = 0.182$
Moles left = 0.4 - 0.182 = 0.218
Molarity $= \frac{0.218 \times 10^{3}}{400} = 0.545 M = 54.5 \times 10^{- 2} M$

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