You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Chemistry

Electrolysis and Electrolytes

A 1.1g sample...

Question

A 1.1g sample of copper ore is dissolved and $C u^{2 +}$ (aq) is titrated with excess $K I$ . The liberated $I_{2}$ requires 12.12 mL of $0.1 M N a_{2} S_{2} O_{3}$ for its titration. What is the $%$ copper by mass in the ore?

Open in App

Solution

$2 C u^{2 +} + 4 I^{-} ⟶ C u_{2} I_{2} + I_{2}$
$I_{2} + 2 S_{2} O_{3}^{2 -} ⟶ S_{4} O_{6}^{2 -} + 2 I^{-}$
Equivalent mass of $C u^{2 +}$ = molar mass
Equivalent mass of $S_{2} O_{3}^{2 -}$ = molar mass
Thus, millimoles of $C u^{2 +}$ = millimoles of $S_{2} O_{3}^{2 -}$
$= 12.12 \times 0.1$
$= 1.212 m i l l i m o l$
$= 1.212 \times 10^{- 3} m o l$
$= 1.212 \times 10^{- 3} \times 63.5 g p u r e C u$
$= 0.076962 g i n 1.1 g C u - o r e$
Thus $%$ of copper $= \frac{0.076962}{1.1} \times 100 = 7%$

Suggest Corrections

Similar questions

Q. A

0.276

g impure sample of copper ore is dissolved and

C u^{2 +}

is titrated with

K I

solution.

I_{2}

liberated required

40

mL of

0.1 M N a_{2} S_{2} O_{3}

solution for titration. The

%

of impurities in the ore is:

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.The amount of

I_{2}

liberated in the reaction of

10

ml of

0.02 M

solution with excess

K I

is :

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.

Percentage of purity of sample is:

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.

The volume (

V

) of hypo required is:

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.

The color of the solution developed during the addition of

K I

C u S O_{4}

solution is :

Join BYJU'S Learning Program

A 1.1g sample of copper ore is dissolved and Cu2+(aq) is titrated with excess KI. The liberated I2 requires 12.12 mL of 0.1 M Na2S2O3 for its titration. What is the % copper by mass in the ore?

A 1.1g sample of copper ore is dissolved and $C u^{2 +}$ (aq) is titrated with excess $K I$ . The liberated $I_{2}$ requires 12.12 mL of $0.1 M N a_{2} S_{2} O_{3}$ for its titration. What is the $%$ copper by mass in the ore?