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Question: The specific conductivity of a saturated solution of \[AgCl\] is \[3.40 \times {10^{ - 6}}{\Omega ^{...

The specific conductivity of a saturated solution of AgClAgCl is 3.40×10−6Ω−1cm−13.40 \times {10^{ - 6}}{\Omega ^{ - 1}}c{m^{ - 1}} at 25∘C{25^ \circ }C.
If λAg+=62.3Ω−1cm2mol−1{\lambda _{A{g^ + }}} = 62.3{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}} and λCl−=67.7Ω−1cm2mol−1{\lambda _{C{l^ - }}} = 67.7{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}} , the solubility of AgClAgCl at 25∘C{25^ \circ }C is:-
A.2.6×10−5molL−12.6 \times {10^{ - 5}}mol{L^{ - 1}}
B.3.731×10−3molL−13.731 \times {10^{ - 3}}mol{L^{ - 1}}
C.3.731×10−5molL−13.731 \times {10^{ - 5}}mol{L^{ - 1}}
D.2.6×10−3gL−12.6 \times {10^{ - 3}}g{L^{ - 1}}

Explanation

Solution

In the above question we are provided with molar conductance of both the ions, which when added will provide the total molar conductance of the salt given that isAgClAgCl. Now to find solubility we will use the formula of molar conductivity.

Complete step by step solution: The question is based on the concept of solubility and molar conductance in a solution. Let’s discuss it in detail.
Solubility in a solution:
Solubility is defined as the ability of a compound which is termed as solute to get dissolve into another substance which is known as a solvent on dissolving one in other we get a solution. When ionic compounds such as salts are dissolved in water or other solvents (which readily dissolves them forms a solution which can conduct electricity due to the cations and anions present in it.
Molar conductivity:
Molar conductivity of a solution is the measurement of how much conducting power does a solution possesses when one mole of electrolyte is dissolved in a solution. It is denoted byλ\lambda , the unit of molar conductivity is Ω−1cm2mol−1{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}}.
Now specific molar conductance is also an important concept, it is denoted by KK and is related to molar conductance λ\lambda as:
λ=K×1000M\lambda = \dfrac{{K \times 1000}}{M} -(1)
Where MM is the molar concentration.
Since in the above question, we have to find out solubility which is same as molar concentration;
We are provided with:
K=3.40×10−6Ω−1cm−1K = 3.40 \times {10^{ - 6}}{\Omega ^{ - 1}}c{m^{ - 1}}
λAg+=62.3Ω−1cm2mol−1{\lambda _{A{g^ + }}} = 62.3{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}}
λCl−=67.7Ω−1cm2mol−1{\lambda _{C{l^ - }}} = 67.7{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}}

λAgCl=λAg++λCl− ⇒λAgCl=(62.3+67.7)Ω−1cm2mol−1 ⇒λAgCl=130Ω−1cm2mol−1  {\lambda _{AgCl}} = {\lambda _{A{g^ + }}} + {\lambda _{C{l^ - }}} \\\ \Rightarrow {\lambda _{AgCl}} = \left( {62.3 + 67.7} \right){\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}} \\\ \Rightarrow {\lambda _{AgCl}} = 130{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}} \\\

Now using the formula in equation 1 and applying all the above data we get
λ=K×1000M\lambda = \dfrac{{K \times 1000}}{M}
By interchanging the subject of the equation to we get;
M=K×1000λAgClM = \dfrac{{K \times 1000}}{{{\lambda _{AgCl}}}}

M=3.4×10−6×1000130 ⇒M=34×10−4130 ⇒M=2.6×10−5molL−1  M = \dfrac{{3.4 \times {{10}^{ - 6}} \times 1000}}{{130}} \\\ \Rightarrow M = \dfrac{{34 \times {{10}^{ - 4}}}}{{130}} \\\ \Rightarrow M = 2.6 \times {10^{ - 5}}mol{L^{ - 1}} \\\

Therefore the answer to the question is option A 2.6×10−5molL−12.6 \times {10^{ - 5}}mol{L^{ - 1}}.

Note: The above-given formula is the most important formula that is being used in the concept of conductivity of the solution, other quantities can be calculated by either interchanging the subject or either substituting another equation with it.