Question

If ${\lambda }_{ClC{H}_{2}COONa}^{\infty }=224{\Omega }^{-1}c{m}^{2}gequi{v}^{-1}$
${\lambda }_{NaCl}^{\infty }=38.5{\Omega }^{-1}c{m}^{2}gequi{v}^{-1}$
${\lambda }_{HCl}^{\infty }=203{\Omega }^{-1}c{m}^{2}gequi{v}^{-1}$
What is the value of ${\lambda }_{ClC{H}_{2}COOH}^{\infty }$?

• A. $288.5{\Omega }^{-1}c{m}^{2}gequi{v}^{-1}$
• B. $289.5{\Omega }^{-1}c{m}^{2}gequi{v}^{-1}$
• C. $388.5{\Omega }^{-1}c{m}^{2}gequi{v}^{-1}$
• D. $59.5{\Omega }^{-1}c{m}^{2}gequi{v}^{-1}$

Answer 1

Answer: (C)

$388.5{\Omega }^{-1}c{m}^{2}gequi{v}^{-1}$

According to Kohlrausch's law
${\lambda }_{ClC{H}_{2}COOH}^{\infty }={\lambda }_{ClC{H}_{2}CO{O}^{-}}^{\infty }+{\lambda }_{H^{+}}^{\infty }$
,
${\lambda }_{ClC{H}_{2}COONa}^{\infty },$

or ${\lambda }_{ClC{H}_{2}CO{O}^{-}}^{\infty }+{\lambda }_{N{a}^{+}}^{\infty }=224$ ..... (i)

${\lambda }_{HCl}^{\infty }={\lambda }_{H^{+}}^{\infty }+{\lambda }_{C{l}^{-}}^{\infty }$

or ${\lambda }_{H^{+}}^{\infty }+{\lambda }_{C{l}^{-}}^{\infty }=203$ ..... (ii)

${\lambda }_{NaCl}^{\infty }={\lambda }_{Na}^{\infty }+{\lambda }_{C{l}^{-}}^{\infty }$

or ${\lambda }_{N{a}^{+}}^{\infty }+{\lambda }_{C{l}^{-}}^{\infty }=38.5$ ..... (iii)

Adding Eqs. (i) and (ii) and subtracting Eq. (iii), we get:

${\lambda }_{C{H}_{3}CO{O}^{-}}^{\infty }+{\lambda }_{H^{+}}^{\infty }$

$=224+203-38.5$

$=427-38.5$

$=388.5{\Omega }^{-1}c{m}^{2}ge{q}^{-1}$

Hence, option C is correct.