Question

Calculate limiting molar conductance for NH4OH, Given molar conductances at infinite dilution for Ba(OH)2, BaCl2 and NH4Cl are 517.6, 240.6 and 129.8 respectively​

Answer 1

Answer:

Did you mean: Calculate limiting molar conductance for NH4OH, Given molar conductances at infinite dilution for Ba(OH)2, CaCl2 and NH4Cl are 517.6, 240.6 and 129.8 respectively

Answer 2

Answer:

Molar conductance at infinite dilution for

$\rm \: NH_4OH$

May be calculated at following process.

$\rm { \Lambda}^{0}_m(NH_4 Cl ) + \dfrac{1}{2} {\Lambda}^{0}_m(Ba(OH)_2) - \dfrac{1}{2} {\Lambda}^{0} _m(BaCl_2)$

$\rm = { \lambda}^{0}_{{NH_{4}}^{ + } } + { \lambda}^{0}_{{Cl_{}}^{ - } } + \frac{1}{2} { \lambda}^{0}_{{Ba_{2}}^{ + } } + { \lambda}^{0}_{{OH_{}}^{ - } } - \frac{1}{2} { \lambda}^{0}_{{Ba_{2}}^{ + } } - { \lambda}^{0}_{{Cl_{}}^{ - } }$

$\rm = { \lambda}^{0}_{{NH_{4}}^{ + } } + \cancel{{ \lambda}^{0}_{{Cl_{}}^{ - } } }+ { \cancel{\frac{1}{2} { \lambda}^{0}_{{Ba_{2}}^{ + } }}}+ { \lambda}^{0}_{{OH_{}}^{ - } } \cancel{ - \frac{1}{2} { \lambda}^{0}_{{Ba_{2}}^{ + } } } - \cancel{{ \lambda}^{0}_{{Cl_{}}^{ - } }}$

$\rm \: = { \lambda}^{0}_{{NH_{4}}^{ + } } + { \lambda}^{0}_{{OH_{}}^{ - } }$

$= \rm{ \Lambda}^{0}NH_4OH$

Given

$\rm {\Lambda}^{0}_m(Ba(OH)_2) = 517.6$

$\rm {\Lambda}^{0} _m(BaCl_2) = 240.6$

$\rm { \Lambda}^{0}_m(NH_4 Cl ) = 129.8$

Calculation:-

$\rm{ \Lambda}^{0}NH_4OH \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = 129.8 + \dfrac{1}{2} \times 517.6 - \dfrac{1}{2} \times 240.6$

$= \rm \: 129.8 + 258.8 - 120.3$

$\underline{\boxed{ \rm =268.3 \: { \: {Ohm}}^{ - 1} {cm}^{2} {mol}^{ - 1} } }$

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