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Electrochemistry

Question
CBSEENCH12011287

Which of the following expression correctly represents the equivalent conductance at infinite dilution of Al2(SO4)3. Given that  straight capital lambda subscript Al to the power of 3 plus end exponent end subscript superscript straight o space and space space straight capital lambda subscript SO subscript 4 superscript 2 minus end superscript end subscript superscript straight oare the equivalent conductance at infinite dilution of the respective ions.

  • straight capital lambda subscript Al to the power of 3 plus end exponent end subscript superscript straight o space plus space space straight capital lambda subscript SO subscript 4 superscript 2 minus end superscript end subscript superscript straight o
  • left parenthesis straight capital lambda subscript Al to the power of 3 plus end exponent end subscript superscript straight o space and space space straight capital lambda subscript SO subscript 4 superscript 2 minus end superscript end subscript superscript straight o right parenthesis space straight x space space 6
  • 1 third space straight capital lambda subscript Al to the power of 3 plus end exponent end subscript superscript straight o space space plus 1 half space space straight capital lambda subscript SO subscript 4 superscript 2 minus end superscript end subscript superscript straight o

Solution

B.

straight capital lambda subscript Al to the power of 3 plus end exponent end subscript superscript straight o space plus space space straight capital lambda subscript SO subscript 4 superscript 2 minus end superscript end subscript superscript straight o

Al2(SO4)3  ⇌ 2 Al3+ + 3SO42-
Since equivalent conductances are given only for ions, the equivalent conductance at infinite dilution,
straight capital lambda subscript eq superscript infinity space equals space straight capital lambda subscript Al to the power of 3 plus end exponent end subscript superscript 0 space plus straight capital lambda subscript SO subscript 4 superscript 2 minus end superscript end subscript superscript straight o