Question

What is the molar conductance of $NaCl$ at infinite dilution?

• A. $147oh{m}^{-1}c{m}^{2}mol{e}^{-1}$
• B. $107oh{m}^{-1}c{m}^{2}smol{e}^{-1}$
• C. $127oh{m}^{-1}c{m}^{2}mol{e}^{-1}$
• D. $157oh{m}^{-1}c{m}^{2}mol{e}^{-1}$

Answer 1

The correct option is C $127oh{m}^{-1}c{m}^{2}mol{e}^{-1}$

${\lambda }_m^c={\lambda }_m^{\infty }-b\sqrt{C}$

When $C_1=4\times {10}^{-4}{\lambda }_m^c=107$ and $C_2=9\times {10}^{-4}{\lambda }_m=97$

$107={\lambda }_m^{\infty }-b\times 2\times {10}^{-2}$ .......... (1)

$97={\lambda }_m^{\infty }-b\times 3\times {10}^{-2}$ .......... (2)

$b=1000$

${\lambda }_m={\lambda }_m^{\infty }-b\sqrt{C}$

${\lambda }_m^{\infty }={\lambda }_m+b\sqrt{C}=107+{10}^3\times 2\times {10}^{-2}$

${\lambda }_m^{\infty }=127oh{m}^{-1}c{m}^{2}mol{e}^{-1}$