JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Some Basic Concepts of Chemistry

When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, _____

\times 1 0^{- 5}

moles of lead sulphate precipitate out.
(Round off to the nearest integer).

Answer: 525

Solution

The lead(II) ion taken is

0.15 \times 0.035 = 5.25 \times 1 0^{- 3}

mol and the chromium(III) sulphate taken is

0.12 \times 0.020 = 2.4 \times 1 0^{- 3}

mol. Each formula unit of

C r_{2} (S O_{4})_{3}

carries three sulphate ions, so the sulphate available is

3 \times 2.4 \times 1 0^{- 3} = 7.2 \times 1 0^{- 3}

mol against only

5.25 \times 1 0^{- 3}

mol of lead:

P b^{2 +} + S O_{4}^{2 -} \to PbS O_{4} ↓

Lead is the limiting reactant, so the precipitate is

5.25 \times 1 0^{- 3}

mol, that is

525 \times 1 0^{- 5}

mol of lead sulphate.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer