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Question

The molar conductivity at infinite dilution of AgNO3, NaCl and NaNO3 are 116.5, 110.3 and 105.2 S cm2mol1 respectively. The conductivity of AgCl in water is 2.40×106Scm1 and of water used is 1.16×106Scm1. The solubility of AgCl will be:

A
1.463×103glitre1
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B
1.163×102glitre1
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C
1.163×103glitre1
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D
None of these
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Solution

The correct option is A 1.463×103glitre1
For monovalent electrolytes,

Equivalent conductivity = Molar conductivity

Given, AAgNO3=116.5=λAg++λNO3 ......(i)

ANaCl=110.3=λNa++λCl ......(ii)

ANaNO3=105.2=λNa++λNO3 ......(iii)

AAgCl=Eqs.[(i)+(ii)(iii)]

=116.5+110.3105.2=121.6
For sparingly soluble salt λv=λ

AvAgCl=121.6Scm2eq1

Now, kAgCl+water=2.40×106Scm1

kwater=1.16×106Scm1

kAgCl=(2.401.16)×106

=1.24×106Scm1

Therefore, Av=k×(1000/M)(N=M)

121.6=1.24×106×(1000/M)

M=1.24×103121.6mol/litre

c=1.24×103121.6×143.5g/litre

Solubility of AgCl =1.463×103glitre1

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