Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1-85- 10 -5 S m -1- Solubility product of the salt AB at 298 K is -Given - m o AB -140- 10 -4 S m 2 mol -1-

The correct option is B 1.74× 10 ^ 12 Λ_m^0=k×1000C 140×10^ 4=1.85×10^ 5×11000C as units are in metres C≃1.32×10^ 6 AB ⇋ A^+ + B^ ...

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Standard XII

Chemistry

Conductance of Electrolytic Solution

Conductivity ...

Question

Conductivity of a saturated solution of a sparingly soluble salt $A B$ at $298 K$ is $1.85 \times 10^{- 5} S m^{- 1}$ . Solubility product of the salt $A B$ at $298 K$ is :

[Given $Λ_{m}^{o} (A B) = 140 \times 10^{- 4} S m^{2} {m o l}^{- 1}$ ]

1.32 \times 10^{- 12}

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1.74 \times 10^{- 12}

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5.7 \times 10^{- 12}

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7.5 \times 10^{- 12}

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Solution

The correct option is B $1.74 \times 10^{- 12}$
$Λ_{m}^{0} = k \times \frac{1000}{C}$

$140 \times 10^{- 4} = 1.85 \times 10^{- 5} \times \frac{1}{1000 C}$ as units are in metres

$C ≃ 1.32 \times 10^{- 6}$

$A B ⇋ A^{+} + B^{-}$

$K_{s p} = C^{2} = 1.74 \times 10^{- 12}$

Hence, option B is correct.

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Conductivity of a saturated solution of a sparingly soluble salt AB at 298K is 1.85×10−5Sm−1. Solubility product of the salt AB at 298K is :[Given Λom(AB)=140×10−4Sm2mol−1]

The correct option is B 1.74×10−12Λ0m=k×1000C140×10−4=1.85×10−5×11000C as units are in metresC≃1.32×10−6AB⇋A++B−Ksp=C2=1.74×10−12Hence, option B is correct.

Conductivity of a saturated solution of a sparingly soluble salt $A B$ at $298 K$ is $1.85 \times 10^{- 5} S m^{- 1}$ . Solubility product of the salt $A B$ at $298 K$ is :

[Given $Λ_{m}^{o} (A B) = 140 \times 10^{- 4} S m^{2} {m o l}^{- 1}$ ]