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  • The solubility of <img alt="\left [ Co\left ( NH_{3} \right )_{4} Cl_{2}\right ]ClO_{4}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cinline%20%5Cleft%20%5B%20Co%5Cleft%20%28%20NH_%7B3%7

The solubility of \left [ Co\left ( NH_{3} \right )_{4} Cl_{2}\right ]ClO_{4}  if the  \lambda _{CO\left ( NH_{3} \right )_{4}Cl_{2}^{+}}=50,\; \; \lambda _{ClO_{4}^{-}}=70     and the measured resistance was 33.5 \Omega in a cell with cell constant of 0.20 is __________in \; mmol/L .

(response should be upto only one decimal digit like 78.7).

Option: 1

59.7


Option: 2

49.7


Option: 3

39.7


Option: 4

29.7


Answers (1)

best_answer

 

Limiting molar conductivity -

When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by \Lambda ^{0}m

-

\\\lambda ^o[Co(NH_3)_4 Cl_2]ClO_4 = \lambda ^o[Co(NH_3)_4 Cl_2] ^+ + \lambda ^oClO_4 ^- \\\lambda ^o[Co(NH_3)_4 Cl_2]ClO_4 = 50+70 =120 \\\lambda ^o[Co(NH_3)_4 Cl_2]ClO_4 = \frac{1000K}{M}

\\K=\frac{l}{RA} \\K=\frac{G^*}{R}= \frac{0.20}{33.5} \\\lambda ^o[Co(NH_3)_4 Cl_2]ClO_4 = 120=\frac{1000\times 0.2}{33.5 \times M} \\M=\frac{1000\times 0.2}{33.5}\times120 = 49.75\ m mol/L

 

Posted by

Irshad Anwar

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