QJEEM23Chc7ee: JEE Main 2023 Chemistry - Class 11 JEEM23Chc7ee Solution: solution of Br is mixed with solution of . The resulting…

Question

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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

Accepted Answer

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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