Question

Equivalent conductivity of $CaC{l}_2$ solution:

• A. $1.6Sc{m}^{2}e{q}^{-1}$
• B. $1.8Sc{m}^{2}e{q}^{-1}$
• C. $2Sc{m}^{2}e{q}^{-1}$
• D. $3Sc{m}^{2}e{q}^{-1}$

Answer 1

The correct option is C $2Sc{m}^{2}e{q}^{-1}$

For KCL solution

$\kappa =\frac{1}{R}\times \frac{l}{a}$

$0.00189=\frac{1}{410.5}\times \frac{l}{a}$

$\frac{l}{a}=0.775845/cm$

For $CaC{l}_2$ solution

$\kappa =\frac{1}{R}\times \frac{l}{a}=\frac{1}{990}\times 0.775845=0.000784$

$N=\frac{11\times 2}{111\times 0.5}=0.396N$

$\lambda =\frac{1000\kappa }{N}=\frac{1000\times 0.000784}{0.396}=1.98\text{S}cm/eq$.