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Q.

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× 10−2M. (Nearest Integer)
(Molar mass of nitric acid is 63 g mol−1 )

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answer is 55.

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Detailed Solution

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Moles of nitric acid  =800×10−3×0.5=0.4
Moles of nitric acid evaporated  =11.563=0.182
Moles left = 0.4 - 0.182 = 0.218
Molarity  =0.218×103400=0.545 M=54.5×10−2M

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