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Question: The specific conductance of a saturated solution of AgCl at \[{25^ \circ }C\] is \[1.821 \times {10^...

The specific conductance of a saturated solution of AgCl at 25∘C{25^ \circ }C is 1.821×10−5mhocm−11.821 \times {10^{ - 5}}mhoc{m^{ - 1}}. What is the solubility of AgCl in water ( in gL−1g{L^{ - 1}}) if limiting molar conductivity of AgCl is 130.26mho cm2 mol−1130.26mho{\text{ }}c{m^2}{\text{ }}mo{l^{ - 1}}?
(A) 1.89×10−3gL−11.89 \times {10^{ - 3}}g{L^{ - 1}}
(B) 2.78×10−2gL−12.78 \times {10^{ - 2}}g{L^{ - 1}}
(C) 2.004×10−2gL−12.004 \times {10^{ - 2}}g{L^{ - 1}}
(D) 1.43×10−3gL−11.43 \times {10^{ - 3}}g{L^{ - 1}}

Explanation

Solution

The relation between specific conductivity of an electrolyte, its molar conductivity and its concentration is given by following formula.
Λm=kc{\Lambda _m} = \dfrac{k}{c}
Solubility of AgCl in will be equal to the concentration we find in the above given formula. Atomic weight of a Silver atom is 108gmmol−1gmmo{l^{ - 1}}.

Complete answer:
Specific conductance(k) of a saturated solution of AgCl is given. Limiting molar conductivity (Λ∘m{\Lambda ^ \circ }_m) of AgCl is also given. So, we can find the concentration of the electrolyte present in the solution which is AgCl in this case. The formula is
Λ∘m=kc{\Lambda ^ \circ }_m = \dfrac{k}{c}..............(1)
But in this formula, k needs to be in mho m1−mho{\text{ }}{{\text{m}}^{1 - }} unit. So, let's convert the given k value.
We know that 1m = 100cm, So,
1.821×10−5mhocm−11.821 \times {10^{ - 5}}mhoc{m^{ - 1}} = 100×1.821×10−5mhom−1100 \times 1.821 \times {10^{ - 5}}mho{m^{ - 1}} = 1.821×10−3mho m−11.821 \times {10^{ - 3}}mho{\text{ }}{m^{ - 1}}
Λ∘m{\Lambda ^ \circ }_m here is also in mho cm2 mol−1mho{\text{ }}c{m^2}{\text{ }}mo{l^{ - 1}} unit which need to be converted to mho m2 mol−1mho{\text{ }}{m^2}{\text{ }}mo{l^{ - 1}}
So, 130.26mho cm2 mol−1130.26mho{\text{ }}c{m^2}{\text{ }}mo{l^{ - 1}} = 0.013026mho m2 mol−10.013026mho{\text{ }}{m^2}{\text{ }}mo{l^{ - 1}}
Also the concentration we will obtain will be in mol m−3mol{\text{ }}{{\text{m}}^{ - 3}} unit that we will convert later. Let’s put available values in eq.(1)
0.013026mho m2 mol−1=1.821×10−3mho m−1c0.013026mho{\text{ }}{m^2}{\text{ }}mo{l^{ - 1}} = \dfrac{{1.821 \times {{10}^{ - 3}}mho{\text{ }}{m^{ - 1}}}}{c}
c=1.821×10−3mho m−10.013026mho m2 mol−1c = \dfrac{{1.821 \times {{10}^{ - 3}}mho{\text{ }}{m^{ - 1}}}}{{0.013026mho{\text{ }}{m^2}{\text{ }}mo{l^{ - 1}}}}
c=0.1398mol m−3c = 0.1398mol{\text{ }}{m^{ - 3}}
Now, the concentration of AgCl able to conduct electricity in the solution is equal to the solubility of AgCl.
So, solubility of AgCl in this case is 0.1398mol m−30.1398mol{\text{ }}{m^{ - 3}} but we are given the answers in gL−1g{L^{ - 1}} values. So, let’s convert them.
Number of moles of the compound = weight of the compoundmolecular weight of the compound{\text{Number of moles of the compound = }}\dfrac{{{\text{weight of the compound}}}}{{{\text{molecular weight of the compound}}}}..........(2)
And Molecular weight of AgCl = Atomic weight of Ag atom + Atomic weight of Cl atom
Molecular weight of AgCl = 108 + 35.5
Molecular weight of AgCl = 143.5gm mol−1gm{\text{ mo}}{{\text{l}}^{ - 1}}
So, we can write equation (2) as
0.1398= weight of AgCl143.50.1398 = \dfrac{{{\text{ weight of AgCl}}}}{{143.5}}
weight of AgCl=0.1398×143.5{\text{weight of AgCl}} = 0.1398 \times 143.5
weight of AgCl=20.06gm{\text{weight of AgCl}} = 20.06gm in 1m−3{m^{ - 3}} volume.
We also know that 1000L=1m−3{m^{ - 3}}
So, 20.06gm m−320.06gm{\text{ }}{m^{ - 3}} = 20.061000gmL−1\dfrac{{20.06}}{{1000}}gm{L^{ - 1}}
So, solubility of AgCl= 2.00×10−2gL−12.00 \times {10^{ - 2}}g{L^{ - 1}}
So, the correct answer is “Option C”.

Note: Remember to put the values of specific conductance and molar conductance in specific required units only. If we are given them in other units, then we need to convert them first. Remember that we can take limiting molar conductivity in place of molar conductivity in the equation shown in Hint part because it is nothing but molar conductivity at a specific condition.