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At 25^(@)C, molar conductivity of 0.007 ...

At `25^(@)C`, molar conductivity of 0.007 M HF acid is `"150 ohm^-1.cm"^(2)."mol"^(-1)`. Its `Lambda_(m)^(@)="500 ohm^-1.cm"^(2)."mol"^(-1)`. The value of the dissociation constant of the acid at the given concentration at `25^(@)C` is -

A

`7xx10^(-4)M`

B

`7xx10^(-5)M`

C

`9xx10^(-3)M`

D

`9xx10^(-4)M`

Text Solution

Verified by Experts

The correct Answer is:
D
`alpha=(Lambda_(m))/(Lambda_(m)^(0))=(150)/(500)=0.3`
`K_(alpha)=(alpha^(2)C)/(1-alpha)=((0.3)^(2)xx0.007)/(1-0.3)=9xx10^(-4)(M)`
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