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A drop of a solution (0.05 mL) contains ...

A drop of a solution `(0.05 mL)` contains `6.0xx10^(-7)` mole of `H^(+)`. If the rate of disappearance of `H^(+)` is `6.0xx10^(5) mol litre^(-1) sec^(-1)`, it takes `x xx10^(-8)` sec for `H^(+)` in the drop to disappear. Find the value of `x`.

Text Solution

Verified by Experts

The correct Answer is:
2
`[H^(+)]=("mole of" H^(+))/("Volume (in L)")`
`=(6.0xx10^(-7))/(0.05xx10^(-3))=1.2xx10^(-2) M`
Now, rate `=(d[H^(+)])/(dt)=6.0xx10^(5)`
`=(1.2xx10^(-2))/(dt)=6.0xx10^(5)`
or `dt=2.0xx10^(-8) sec`
`:. x=2`
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