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Solubility Product

The specific ...

Question

The specific conductance of a saturated solution of AgCl at $25^{\circ} C$ after substracting the specific conductance of water is $2.28 \times 10^{- 6} m h o c m^{- 1}$ . The solubility product of AgCl at $25^{\circ} C$ is ( $λ_{A g C l}^{\infty} = 138.3 m h o c m^{2}$ )

3.70 \times 10^{- 10} (m o l e / l i t r e)^{2}

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1.70 \times 10^{- 10} (m o l e / l i t r e)^{2}

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0.70 \times 10^{- 10} (m o l e / l i t r e)^{2}

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N o n e o f t h e s e

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Solution

The correct option is D $N o n e o f t h e s e$
$2.70 \times 10^{- 10} (m o l e / l i t r e)^{2}$

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Similar questions

A g C l

25^{\circ} C

3.55 \times 10^{- 6} Ω^{- 1} c m^{- 1}

1.60 \times 10^{- 6} Ω^{- 1} c m^{- 1}

(Λ^{\infty} (A g C l) = 150.0 Ω^{- 1} c m^{2} {eqv}^{- 1})

A g C l

25^{\circ} C

3.55 \times 10^{- 6} Ω^{- 1} c m^{- 1}

1.60 \times 10^{- 6} Ω^{- 1} c m^{- 1}

(Λ^{\infty} (A g C l) = 150.0 Ω^{- 1} c m^{2} {eqv}^{- 1})

25^{o} C

2.28 \times 10^{- 4} S m^{- 1}

d m^{3}

$Λ_{m}^{0} (A g C l) = 138.3 \times 10^{- 4} S m^{2} m o l^{- 1}$
$M (A g C l) = 143.5 g m o l^{- 1}$

25^{0}

10^{- 4} S m^{- 1}

d m^{3}

A g^{+}

C l^{-}

10^{- 4}

10^{- 4} S m^{2} m o l^{- 1}

1.826 \times 10^{- 6} Ω^{- 1} c m^{- 1}

25^{\circ} C

25^{\circ} C

λ^{\circ} (A g^{+}) = 61.92

λ^{\circ} (C l^{-}) = 76.34

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