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Standard XII

Chemistry

Kohlrausch Law

The specific ...

Question

The specific conductance of a saturated solution of AgCl is “ $κ$ ” $Ω^{- 1} c m^{- 1}$ . The limiting ionic conductance of $A g^{+}$ and $C l^{-}$ ions are x and y respectively. Then, the solubility product of AgCl is:

\frac{100 κ}{(x + y)}

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\frac{(1000 κ)^{2}}{(x + y)^{2}}

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\frac{(1000 \times 14.3 \times κ)}{(x + y)}

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\frac{(10^{3} \times 143.5 \times κ)^{2}}{(x + y)^{2}}

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Solution

The correct option is B $\frac{(1000 κ)^{2}}{(x + y)^{2}}$
Point to remember:
$\Rightarrow S o l u b l i t y p r o d u c t :$
$A B S \to A^{+} S + B^{-} S$
$K_{s p} = S^{2}$
$\Rightarrow λ_{m}^{0} = 1000 κ / C$
$C = S$
given data:
$λ_{A g^{+}}^{0} = x$
$λ_{C l^{-}}^{0} = y$
so $λ_{A g C l}^{0} = x + y$
On substituting we get
$\Rightarrow λ_{A g C l}^{0} = 1000 κ / S$
$S = \frac{1000 κ}{x + y}$
So $K_{s p} = (\frac{1000 κ}{x + y})^{2}$

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Similar questions

The specific conductance of a saturated solution of silver chloride is $K Ω^{- 1} c m^{- 1}$ . The limiting ionic conductances of $A g$ $^{+}$ and $C l$ $^{-}$ ions are $x$ and $y$ , respectively. The solubility of silver chloride in g litre $^{- 1}$ is :

[Molar mass of $A g C l = 143.5$ ]

Q. The specific conductance of a saturated solution of

A g C l

K Ω^{- 1} c m^{- 1}

. The amount of

A g

and

C l^{-}

are

x

and

y

respectively. The solubility product of

A g C l

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Q. The specific conductance of a saturated solution of AgCl at

25^{0}

C after subtracting the specific conductance of water is 2.28 x

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

. Calculate the solubility of AgCl in gram per

d m^{3}

at this temperature. The molar ionic conductances of

A g^{+}

and

C l^{-}

ions are 73.3 x

10^{- 4}

and 65.0 x

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

. (Ag: 108 and Cl: 35.5).

Q. The specific conductance of

A g C l

at 298 K is

2.326 \times 10^{- 6} o h m^{- 1} c m^{- 1}

. The ionic conductivities of

A g^{+}

and

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ions are

71.92

and

86.34

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respectively. What is the solubility of

A g C l

in water? (Molarmass of

A g C l

143.5 g m o l^{- 1}

)

The specific conductance of $A g C l$ at 298 K is $1.826 \times 10^{- 6} o h m^{- 1} c m^{- 1}$ . The ionic conductivities of $A g^{+}$ and $C l^{-}$ ions are $61.92$ and $76.34$ $o h m^{- 1} c m^{2} m o l^{- 1}$ respectively. What is the solubility of $A g C l$ in water? (Molar mass of $A g C l$ is $143.5 g m o l^{- 1}$ )

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The specific conductance of a saturated solution of AgCl is “κ” Ω−1cm−1. The limiting ionic conductance of Ag+ and Cl− ions are x and y respectively. Then, the solubility product of AgCl is:

The correct option is B (1000κ)2(x+y)2 Point to remember: ⇒Solublity product: ABS→A+S+B−S Ksp=S2 ⇒λ0m=1000κ/C C=S given data: λ0Ag+=x λ0Cl−=y so λ0AgCl=x+y On substituting we get ⇒λ0AgCl=1000κ/S S=1000κx+y So Ksp=(1000κx+y)2

The specific conductance of a saturated solution of AgCl is “ $κ$ ” $Ω^{- 1} c m^{- 1}$ . The limiting ionic conductance of $A g^{+}$ and $C l^{-}$ ions are x and y respectively. Then, the solubility product of AgCl is: