Following data is given for the reaction,$\mathrm{X}+\mathrm{Y}\to \mathrm{P}$

Experiment	$\left[\mathrm{X}\right] \mathrm{mol} {\mathrm{L}}^{-1}$	$\left[\mathrm{Y}\right] \mathrm{mol} {\mathrm{L}}^{-1}$	Initial rate $\left(\mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{s}}^{-1}\right)$
I	$2.5\times {10}^{-4}$	$6\times {10}^{-5}$	$5\times {10}^{-4}$
II	$5\times {10}^{-4}$	$12\times {10}^{-5}$	$4\times {10}^{-3}$
III	$1\times {10}^{-3}$	$12\times {10}^{-5}$	$1.6\times {10}^{-2}$

The order of reaction with respect to X and Y respectively are

Let the order of reaction with respect ta X and Y be a and b, the rate law expression $\mathrm{r}=\mathrm{k}\left[\mathrm{X}{]}^{\mathrm{a}}\right[\mathrm{Y}{]}^{\mathrm{b}}$
Similarly, $\text{ Exp. I. }5\times {10}^{-4}=\mathrm{k}{\left[2.5\times {10}^{-4}\right]}^{\mathrm{a}}{\left[6\times {10}^{-5}\right]}^{\mathrm{b}}$            ………(i)
                $\text{ Exp. II. }4\times {10}^{-3}=\mathrm{k}{\left[5\times {10}^{-4}\right]}^{\mathrm{a}}{\left[12\times {10}^{-5}\right]}^{\mathrm{b}}$            ………(ii)
                $\text{ Exp. III. }1.6\times {10}^{-2}=\mathrm{k}{\left[1\times {10}^{-3}\right]}^{\mathrm{a}}{\left[12\times {10}^{-5}\right]}^{\mathrm{b}}$       ………(iii)
From Eqs. (i) and (ii),
            $\left(\frac{4\times {10}^{-3}}{5\times {10}^{-4}}\right)={\left(\frac{5\times {10}^{-4}}{2.5\times {10}^{-4}}\right)}^{\mathrm{a}}{\left(\frac{12\times {10}^{-5}}{6\times {10}^{-5}}\right)}^{\mathrm{b}}$
                                           $8=\left(2{)}^{\mathrm{a}}\right(2{)}^{\mathrm{b}}$                                         ……..(iv)
From Eq. (ii) and (iii),
            $\left(\frac{1.6\times {10}^{-2}}{4\times {10}^{-3}}\right)={\left(\frac{1\times {10}^{-3}}{5\times {10}^{-4}}\right)}^{\mathrm{a}}{\left(\frac{12\times {10}^{-5}}{12\times {10}^{-5}}\right)}^{\mathrm{b}}$
                             $\begin{array}{r}4=(2{)}^{\mathrm{a}}\\ (2{)}^2=(2{)}^{\mathrm{a}}\end{array}$
                          $\therefore \mathrm{a}=2$       
On substituting value of a in Eq. (iv), we get 
                                   $8=\left(2{)}^2\right(2{)}^{\mathrm{b}}$
                                 $\left(2\right)=(2{)}^{\mathrm{b}}\text{ or }\mathrm{b}=1$
Thus, order of reaction with respect ta X is 2 and with respect to Y is 1.