Question

The equivalent conductivity conductivity of 1M ${\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$ solution would be if specific conductance is $\text{26}\times \text{1}{\text{0}}^{-2}\text{S}\text{c}{\text{m}}^{-1}$.

• A. $1.3\times {10}^2\text{S}\text{c}{\text{m}}^2\text{e}{\text{q}}^{-1}$
• B. $1.6\times {10}^2\text{S}\text{c}{\text{m}}^2$
• C. $13\text{S}\text{c}{\text{m}}^2\text{mo}{\text{l}}^{\text{- 1}}$
• D. $1.3\times {10}^3\text{S}\text{c}{\text{m}}^2\text{mo}{\text{l}}^{-1}$

Answer 1

The correct option is A $1.3\times {10}^2\text{S}\text{c}{\text{m}}^2\text{e}{\text{q}}^{-1}$

Normality = Molarity $\times 2\text{factor}$

$=1\times 2=2\text{N}$ $2{\text{H}}^{+}+S\text{c}{\text{m}}^2$

$\Delta eq=\frac{\kappa \times 1000}{N}=\frac{2.6\times {10}^{-2}\times 1000Sc{m}^{-12=2}}{2}$ $1Lt={10}^{3}c{m}^3$

$=1.3\times 10/{10}^{3}c{m}^3$

$=1.3\times {10}^{-1+3}Sc{m}^2=1.3\times {10}^{2}Sc{m}^{2}e{q}^{-1}$