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Answer :

`8.66 xx 10^(-9) mol L^(-1) s^(-1)`Solution :

`r=k[X][Y]^2 and k= 3.0 xx 10^(-6) mol^(-2) L^2 s^(-1)` <br> Concentration of X after 0.02 mole has been reacted, [X] =0.1-0.2=0.08 mol `L^(-1)` <br> From the given reaction, it is clear that when 2 moles of X are consumed then one mole of Y gets consumed. <br> `therefore ` .. Moles of Y consumed when 0.02 moles of X have been consumed `= 0.02 xx 1/2 = 0.01 `mole <br> ` therefore [Y] = 0.2-0.01 = 0.19 mol L^(-1)` <br> Rate ` = ( 3.0 xx 10^(-6)) mol^(-2) L^2 s^(-1) ( 0.08 mol L^(-1) ) ( 0.019 mol L^(-1))^2` <br> ` = 8.66 xx 10^(-9) mol L^(-1) s^(-1)`