The solubility product of Zn(OH)₂ at 298K is 1.2 × 10^-17. The molar ionic conductances of ${Zn}^{2 +}, {Na}^{+}, {OH}^{-} and {Cl}^{-}$ ions are 106.0, 50.0, 200.0 and 76.0 ohm^-1cm² mol^-1
respectively. Incorrect statement(s) is (are)

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$The solubility of Zn (OH)_{2} is (in mol d m^{- 3}) 1.44 \times 10^{- 4}$

$The molar conductance of Zn (OH)_{2} at infinite dilution is (in {ohm}^{- 1} {cm}^{2} {mol}^{- 1}) 506 .0$

$The mobility of {Zn}^{2 +} is (in {cm}^{2} {volt}^{- 1} \sec^{- 1}) 5.5 \times 10^{- 3} .$

The specific conductance of a saturated solution of Zn(OH)₂ at this temperature $is (in {ohm}^{- 1} {cm}^{- 1}) 7.28 \times 10^{- 7}$

answer is A, D.

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Detailed Solution

$\begin{aligned} (1) K_{SP} = 4 S^{3} = 1 .2 \times 10^{- 17}; S^{3} = 3 \times 10^{- 18} \\ S = \sqrt[3]{3 \times 10^{- 18}} = 1 .44 \times 10^{- 6} mol {dm}^{- 3} \end{aligned} \begin{aligned} (2) μ_{\infty} (Zn (OH)_{2}) = μ_{\infty} ({ZnCl}_{2}) + 2 μ_{\infty} (NaOH) - 2 μ_{\infty} (NaCl) \\ = (106.0 + 2 \times 76.0) + 2 (50 + 200) - 2 (50 + 76) \\ = 258.0 + 500 - 252 = 506.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned} (3) o r S = \frac{1000 k}{Λ_{\infty}} = \frac{1000 k}{506} ∴ k = \frac{1.44 \times 10^{- 6} \times 506}{1000} = 7.28 \times 10^{- 7} s {cm}^{- 1} (4) μ_{{Zn}^{2} +}^{0} = \frac{106}{2 \times 96, 500} = 5.5 \times 10^{- 4} {cm}^{2} {volt}^{- 1} \sec^{- 1}$