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Characteristics of Equilibrium Constant

Mixture [X]...

Question

Mixture $[X] = 0.02$ mole of $[C o (N H_{3})_{5} S O_{4}] B r$ and 0.02 mol of $[C o (N H_{3})_{5} B r] S O_{4}$ was prepared in 2 litre of solution.
1 litre of mixture [X] + excess $A g N O_{3} ⟶$ [Y]
1 litre of mixture [X] + excess $B a C l_{2} ⟶$ [Z]
No. of moles of [Y] and [Z] are :

A

0.01, 0.01

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B

0.02, 0.01

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C

0.01, 0.02

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D

0.02, 0.02

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Solution

The correct option is A $0.01, 0.01$
Total volume of the solution is 2 L,

Mixture X contains 0.02 moles of $[C o (N H_{3})_{5} S O_{4}] B r$

Therefore Molarity of $[C o (N H_{3})_{5} S O_{4}] B r,$

$\frac{No. of moles of solute}{Volume of solution in Litres} = \frac{0.02}{2} = 0.01 M$

Mixture of X also contains 0.02 moles of $[C o (N H_{3})_{5} B r] S O_{4},$

Therefore, Molarity of $[C o (N H_{3})_{5} B r] S O_{4} =$ $\frac{No. of moles of solute}{Volume of solution in Litres} = \frac{0.02}{2} = 0.01 M$

Now, according to the reaction,

$[C o (N H_{3})_{5} S O_{4}] B r (0.01 M) + A g N O_{3} \to [C o (N H_{3})_{5} S O_{4}] N O_{3} (S o l u b l e) + A g B r [Y] p p t ↓$

According to Avogadro law, the molarity of $A g B r$ will also be 0.01, since molarity is moles per litre. Therefore, the number of moles of $A g B r$ will also be 0.01.

Similarly in the second reaction, $B a S O_{4}$ (Z) will be precipitated and $[C o (N H_{3})_{5} B r] C l_{2}$ will be formed which is soluble. Therefore, no. of moles of $B a S O_{4}$ = 0.01

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Similar questions

X

0.02

[C o (N H 3)_{5} S O_{4}] B r

[C o (N H 3)_{5} B r] S O_{4}

was prepared in 2 litre of solution.
1 litre of mixture

X +

A g N O 3 \to

Y.
1 litre of mixture

X +

B a C 1_{2} \to Z

Number of moles of

Y

Z

Q. Mixture X of 0.02 mole of

[C o (N H_{3})_{4} S O_{4}] B r

and 0.02 mole of

[C o (N H_{3})_{5} B r] S O_{4}

was prepared in 2 litre of solution:

1 litre of mixture X + excess of

A g N O_{3} \to Y

1 litre of mixture X + excess of

B a C l_{2} \to Z

Number of moles of Y and Z respectively are :

X = 0.02

[C o (N H_{3})_{5} S O_{4}] B r

and 0.02 mol of

[C o (N H_{3})_{5} B r] S O_{4}

was prepared in 2L of solution.

1L of mixture X + excess

A g N O_{3} \to Y

1L of mixture X + excess

B a C l_{2} \to Z

Number of moles of Y and Z are :

X

0.02

[C o (N H_{3})_{5} S O_{4}] B r

0.02

[C o (N H_{3})_{5} B r] S O_{4}

was prepared in

2

1

X

A g N O_{3}

\to Y

1

X

B a C l_{2}

\to Z

Number of moles of

Y

Z

is, respectively :

X = 0.02 m o l

[C o (N H_{3})_{5} S O_{4}] B r

0.02 m o l

[C o (N H_{3})_{5} B r] S O_{4}

was prepared in 2 litre of solution.

1 litre of mixture X+ excess A g N O_{3} \to Y

1 litre of mixture X + excess B a C l_{2} \to Z

Number of moles of Y and Z are

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