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Question: The specific conductance of a saturated solution of AgCl at \(25ºC\) is \(1.821 \times 10^{- 5}mhocm...

The specific conductance of a saturated solution of AgCl at 25ºC25ºC is 1.821×10−5mhocm−11.821 \times 10^{- 5}mhocm^{- 1}. What is the solubility of AgCl in water (in gL−1gL^{- 1}), if limiting molar conductivity of AgCl is 130.26mhocm2mol−1?130.26mhocm^{2}mol^{- 1}?

A

1.89×10−3gL−11.89 \times 10^{- 3}gL^{- 1}

B

2.78×10−2gL−12.78 \times 10^{- 2}gL^{- 1}

C

2.004×10−2gL−12.004 \times 10^{- 2}gL^{- 1}

D

1.43×10−3gL−11.43 \times 10^{- 3}gL^{- 1}

Answer

2.004×10−2gL−12.004 \times 10^{- 2}gL^{- 1}

Explanation

Solution

Solubility =k×1000Δºm= \frac{k \times 1000}{\Delta º_{m}}

=1.821×10−5×1000130.26=13.97×10−5molL−1= \frac{1.821 \times 10^{- 5} \times 1000}{130.26} = 13.97 \times 10^{- 5}molL^{- 1}

=13.97×10−5×143.5= 13.97 \times 10^{- 5} \times 143.5 (AgCl=108+35.5=143.5)(AgCl = 108 + 35.5 = 143.5)

=2.004×10−2gL−1= 2.004 \times 10^{- 2}gL^{- 1}