1.

A gaseous hypothetical chemical equations ,is carried out in a closed vessel . The concentration of B is found to increase by 5 xx 10^(-3) mol l^(-1) in 10 second . The rate of appearance of B is

Answer»

`5 xx 10^(-4) mol L^(-1) sec^(-1)`
`5 xx 10^(-5) mol l^(-1) sec^(-1)`
`6 xx 10^(-5) mol l^(-1) sec^(-1)`
`4 xx 10^(-4) mol l^(-1) sec^(-1)`

Solution :Increase in CONCENTRATION of `B = 5 xx 10^(-3)` mol`l^(-1)` Time = 50 sec
Rate of apperance ofB = `("Increase of conc. B")/("Time TAKEN")`
`= (5xx 10^(-3) mol l^(-1))/(10"sec") = 5 xx 10^(-4) mol l^(-1) sec^(-1)`.


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