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Chemistry Question on Electrochemistry

The specific conductance of a saturated solution of AgClAgCl at 25 ∘C25\, ^{\circ}C is 1.821×10−5 mho cm−11.821 \times 10^{-5}\, mho\, cm^{-1}. What is the solubility of AgClAgCl in water (in g L−1g \,L^{-1}), if limiting molar conductivity of AgClAgCl is 130.26 mho cm2 mol−1130.26\, mho\, cm^2\, mol^{-1} ?

A

1.89×10−3 g L−11.89 \times 10^{-3}\,g\,L^{-1}

B

2.78×10−2 g L−12.78 \times 10^{-2}\,g\,L^{-1}

C

2.004×10−2 g L−12.004\times10^{-2}\,g\,L^{-1}

D

1.43×10−3 g L−11.43 \times 10^{-3}\,g\,L^{-1}

Answer

2.004×10−2 g L−12.004\times10^{-2}\,g\,L^{-1}

Explanation

Solution

Solubility =κ×1000Λm∘=\frac{\kappa\times1000}{\Lambda^{\circ}_{m}} =1.821×10−5×1000130.26=\frac{1.821\times10^{-5}\times1000}{130.26} =13.97×10−5 mol L−1=13.97\times10^{-5}\,mol\,L^{-1} =13.97×10−5×143.5(AgCl=108+35.5=143.5)=13.97\times10^{-5}\times143.5\quad\left(AgCl=108+35.5=143.5\right) =2.004×10−2 g L−1=2.004\times10^{-2}\,g\,L^{-1}