A first order reaction is 50% completed after 30 minutes.This implies that the time required to complete 90% of the reaction is?

The integrated rate law for a first-order reaction looks like this

#color(blue)(ln( A/A_0) = - k * t)" "#, where#

#A# - the concentration at a given time #t#
#A_0# - the initial concentration
#k# - the rate constant, usually expressed in #"s"^(-1)# for first-order reactions

Now, I'll assume that you're not familiar with the equation that describes the half-life of a first-order reaction.

As you know, the half-life of a chemical reaction tells you how much time is needed for the concentration of a reactant to reach half of its initial value.

Simply put, the time needed for #50%# of a reaction to be completed will represent that reaction's half-life.

In your case, you know that your first-order reaction is #50%# completed after #30# minutes. This means that after #30# minutes, you will have

#A = 1/2 * A_0 -># half of the initial concentration remains after one half-life

Plug this into the above equation and solve for #k#, the rate constant

#ln( (1/2 * color(red)(cancel(color(black)(A_0))))/(color(red)(cancel(color(black)(A_0)))) ) = - k * "30 min"#

This is equivalent to

#- k* "30 min" = ln(1/2)#

#-k * "30 min" = overbrace(ln(1))^(color(blue)(=0)) - ln(2)#

#k = ln(2)/"30 min" = 2.31 * 10^(-2)"min"^(-1)#

Now that you know the rate constant for this reaction, use the same equation again, only this time use

#A = 1/10 * A_0 -># equivalent to #90%# completion

and solve for #t#, the time needed to get here

#ln( (1/10 * color(red)(cancel(color(black)(A_0))))/(color(red)(cancel(color(black)(A_0)))) ) = - 2.31 * 10^(-2)"min"^(-1) * t#

This is equivalent to

#t = (color(red)(cancel(color(black)(-))) ln(10))/(color(red)(cancel(color(black)(-))) 2.31 * 10^(-2)"min"^(-1)) = "99.68 min"#

Rounded to one sig fig, the answer will be

#t = color(green)("100 min")#