1.

The half-life of a first order reaction is 10 minutes. If initial amount is 0.08 mol/L and concentration at some instant is 0.01 mol/L, then t is

Answer»

10 minutes
30 minutes
20 minutes
40 minutes.

Solution :`K=(0.693)/(t_(1//2))=(0.693)/(10)`
`t=(2.303)/(k)"LOG"(0.08)/(0.01)=(2.303xx10)/(0.693)log` 8=30 minutes


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