Explanation

(a)
Mixture $X$ contains 0.02 moles of$\ \lbrack Co\left( NH_{3} \right)_{5}SO_{4}\rbrack Br\$and 0.02 moles$\$of$\ \left\lbrack Co\left( NH_{3} \right)_{5}Br \right\rbrack SO_{4}\$was prepared in 2L of solution. So, the concentration of $\lbrack Co\left( NH_{3} \right)_{5}SO_{4}\rbrack Br\ and\ \lbrack Co\left( NH_{3} \right)_{5}Br\rbrack SO_{4}\$in solution are 0.01 mol/L and 0.01 mol/L respectively. During the reaction with$\ AgNO_{3}$(excess), AgBr is precipitated as follows
$\left\lbrack Co\left( NH_{3} \right)_{5}\ SO_{4} \right\rbrack Br + AgNO_{3} \rightarrow \left\lbrack Co\left( NH_{3} \right)_{5}SO_{4} \right\rbrack NO_{3} + AgBr \downarrow$
0.01 mol/L (excess) soluble $(Y)$ 0.01 mol/L
Hence, number of moles of $y$=0.01
On addition of excess $BaCl_{2},BaSO_{4}$ is precipitated as follows
$\left\lbrack Co\left( NH_{3} \right)_{5}Br \right\rbrack SO_{4} + BaCl_{2} \rightarrow BaSO_{4} \downarrow + \left\lbrack Co\left( NH_{3} \right)_{5}Br \right\rbrack Cl_{2}$
0.01 mol/L (excess) 0.01 mol/L soluble ‘$Z$’
Hence, number of moles of $Z =$0.01
Thus, the number of moles of $Y$ and $Z$ are 0.01 and 0.01 respectively.