JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Some Basic Concepts of Chemistry

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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 1 0^{- 2}

M.
(Nearest Integer)
(Molar mass of nitric acid is

63 g mol^{- 1}

)

Answer: 54

Solution

Initial moles of

HNO X_{3}

n = M \times V = 0.5 \times 0.8 = 0.4

mol.

Moles evaporated:

\frac{11.5}{63} = 0.1825

mol.

Remaining moles:

0.4 - 0.1825 = 0.2175

mol.

Final volume:

\frac{800}{2} = 400

mL = 0.4 L.

Molarity:

M = \frac{0.2175}{0.4} = 0.5437

= 54.37 \times 1 0^{- 2}

M.

Nearest integer: 54.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer

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