The volume of water needed to dissolve 1 mg of PbSO4 Ksp - 1-44 - 10-8 M2 at 25 -C is approximately-molar mass of Pb - 207 g-mol

Solubility of PbSO_4 = nbsp;√(K_sp) = √(1.44 × 10^ 8) = 1.2 × 10^ 4 M Solubility of PbSO_4 = 1.2 × 10^ 4 nbsp;× 303 × 10^3 = 36.36 mg L^ 1 ∴Volume of water ...

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Byju's Answer

Standard XII

Chemistry

Solubility Product

The volume of...

Question

The volume of water needed to dissolve $1 mg$ of $P b S O_{4} (K_{s p} = 1.44 \times 10^{- 8} M^{2})$ at $25^{\circ} C$ is approximately:
(molar mass of $P b = 207 g/mol$ )

10 mL

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27 mL

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43 mL

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80 mL

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Solution

The correct option is B $27 mL$
$Solubility of P b S O_{4} = \sqrt{K_{s p}} = \sqrt{1.44 \times 10^{- 8}} = 1.2 \times 10^{- 4} M Solubility of P b S O_{4} = 1.2 \times 10^{- 4} \times 303 \times 10^{3} = 36.36 {mg L}^{- 1} ∴ Volume of water needed to dissolve 1 mg of P b S O_{4} = \frac{1000}{36.36} = 27.5 mL$

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The volume of water needed to dissolve 1 mg of PbSO4(Ksp=1.44×10−8M2) at 25 ∘C is approximately: (molar mass of Pb=207 g/mol)

The correct option is B 27 mLSolubility of PbSO4=√Ksp=√1.44×10−8=1.2×10−4 MSolubility of PbSO4=1.2×10−4×303×103=36.36 mg L−1∴Volume of water needed to dissolve 1 mg of PbSO4=100036.36=27.5 mL

The volume of water needed to dissolve $1 mg$ of $P b S O_{4} (K_{s p} = 1.44 \times 10^{- 8} M^{2})$ at $25^{\circ} C$ is approximately:
(molar mass of $P b = 207 g/mol$ )