(ch3)3cbr +oh- ...(ch3)3coh+ br-

what would be the rate law

The reactant in the reaction is tertiary alkyl halide. It will follow the SN1 mechanism because of the formation of stable carbocation (tertiary carbocation is stable). It is a first order reaction as in SN‚Äč1 mechanism , the rate of reaction only depends on the concentration of the alkyl halide . Rate does not depends on the concentration of the nucleophile. 
The equation is:
(CH3)3CBr(aq) + OH-(aq) → (CH3)3COH(aq) + Br-
The above reaction occurs in two steps:
(CH3)3CBr → (CH3)3C+ Br-   (slow)
(CH3)3C+ OH- → (CH3)3COH  (fast)
The slowest step is the rate determining step. Hence, the rate law is written as :
Rate = k[(CH3)3CBr)]
Hence , it is a first order reaction.

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