The conductance of a $$0.0015$$ M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrode is $$120$$ cm with an area of cross section of $$1$$ $$cm2$$. The conductance of this solution was found to be $$5$$×$$10$$−$$7S$$. The pH of the solution is $$4$$. The value of limiting molar conductivity Λ$$m0$$ of this weak monobasic acid in aqueous solution is Z×$$102S$$ cm−$$1$$ mol−$$1$$ . The value of Z is

Question

The conductance of a $$0.0015$$ M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrode is $$120$$ cm with an area of cross section of $$1$$ $$cm2$$. The conductance of this solution was found to be $$5$$×$$10$$−$$7S$$. The pH of the solution is $$4$$. The value of limiting molar conductivity Λ$$m0$$ of this weak monobasic acid in aqueous solution is Z×$$102S$$ cm−$$1$$ mol−$$1$$ . The value of Z is

Infinity Learn · Accepted Answer

$$K=conductance$$× cell constant $$k=5$$×$$10$$−$$7S$$×$$1201cmcm2=600$$×$$10$$−$$7Scm$$−$$1$$Λ$$m=600$$×$$10$$−$$7$$×$$1000C=600$$×$$10$$−$$7$$×$$10000.0015M=40pH=4(H+)=C$$α$$=10$$−$$4$$ΛMΛ$$m0=$$αΛ$$m0=$$Λmα$$=401/15=600600S$$−$$cm2mol$$−$$12$$×$$102=6$$×$$102$$

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