Question

The molar conductivities are zero concentrations of $N{H}_{4}Cl$,NaOH and NaCl are respectively $149.7{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$, $248.1{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$ and $126.5{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$. What is the molar conductivity of $N{H}_{4}OH$ at zero concentration ?

• A. $261.3{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
• B. $271.3{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
• C. $281.3{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
• D. $291.3{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is B $271.3{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }N{H}_{4}OH=?$

${\Lambda }_m^{\infty }\left(N{H}_{4}Cl\right)=149.7{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }\left(NaOH\right)=248.1{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }\left(NaCl\right)=126.5{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }\left(N{H}_{4}Cl\right)={\Lambda }_{N{H}_4^{+}}^{\infty }+{\Lambda }_{C{l}^{-}}^{\infty }=149.7$.....(i)

${\Lambda }_m^{\infty }NaOH={\Lambda }_{N{a}^{+}}^{\infty }+{\Lambda }_{O{H}^{-}}^{\infty }=248.1$.....(ii)

${\Lambda }_m^{\infty }NaCl={\Lambda }_{N{a}^{+}}^{\infty }+{\Lambda }_{C{l}^{-}}^{\infty }=126.5$.....(iii)

Add equation (i) and (ii) and subtract equation (iii)

${\Lambda }_m^{\infty }N{H}_{4}OH={\Lambda }_{N{H}_{4}Cl}^{\infty }+{\Lambda }_{NaOH}^{\infty }-{\Lambda }_{NaCl}^{\infty }$

$={\Lambda }_{N{H}_4^{+}}^{\infty }+{\Lambda }_{O{H}^{-}}^{\infty }$

$=149.7+248.1-126.5$

${\Lambda }_m^{\infty }N{H}_{4}OH=271.3{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$