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Solubility product of Ba(OH)2 is 4xx10^-...

Solubility product of `Ba(OH)_2` is `4xx10^-9` its solubbility in water is

A

`1xx10^(-3)` M

B

`1xx10^(-9)` M

C

`4xx10^(-27)` M

D

`1xx10^(-27)` M

Text Solution

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The correct Answer is:
A
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