JEE MAIN - Chemistry (2023 - 30th January Evening Shift - No. 13)
$1 \mathrm{~L}, 0.02 \mathrm{M}$ solution of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right]$ Br is mixed with $1 \mathrm{~L}, 0.02 \mathrm{M}$ solution of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}$. The resulting solution is divided into two equal parts $(\mathrm{X})$ and treated with excess of $\mathrm{AgNO}_{3}$ solution and $\mathrm{BaCl}_{2}$ solution respectively as shown below:
$1 \mathrm{~L}$ Solution $(\mathrm{X})+\mathrm{AgNO}_{3}$ solution (excess) $\longrightarrow \mathrm{Y}$
$1 \mathrm{~L}$ Solution $(\mathrm{X})+\mathrm{BaCl}_{2}$ solution (excess) $\longrightarrow \mathrm{Z}$
The number of moles of $\mathrm{Y}$ and $\mathrm{Z}$ respectively are
$1 \mathrm{~L}$ Solution $(\mathrm{X})+\mathrm{AgNO}_{3}$ solution (excess) $\longrightarrow \mathrm{Y}$
$1 \mathrm{~L}$ Solution $(\mathrm{X})+\mathrm{BaCl}_{2}$ solution (excess) $\longrightarrow \mathrm{Z}$
The number of moles of $\mathrm{Y}$ and $\mathrm{Z}$ respectively are
$0 .01,0.01$
$0.01,0.02$
$0.02,0.01$
$0.02,0.02$
Explanation
On mixing both $$\mathrm{\left[ {Co{{(N{H_5})}_5}S{O_4}} \right]Br}$$ and $$\mathrm{\left[ {Co{{(N{H_3})}_5}Br} \right]S{O_4}}$$ becomes 0.01 molar.
$$\therefore$$ Moles of y and z formed will also be 0.01 both.
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