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

Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

I2 obtained f...

Question

$I_{2}$ obtained from $0.1$ mol of $C u S O_{4}$ required $100$ mL of $1$ M hypo solution. Hence, mole percentage of pure $C u S O_{4}$ is:

100

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

50

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $100$
The reactions that occur are given below:
$2 C u S O_{4} + 4 K I \to C u_{2} I_{2} + I_{2} + 2 K_{2} S O_{4}$
$I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}$
$2 C u S O_{4} \equiv 2 N a_{2} S_{2} O_{3}$
$C u S O_{4} \equiv N a_{2} S_{2} O_{3}$
$100$ mL of $1$ M hypo $= 0.1$ mole hypo $\equiv 0.1$ mol pure $C u S O_{4}$
Hence, it is $100 %$ pure.

Suggest Corrections

Similar questions

638.0

g of

C u S O_{4}

solution is titrated with excess of

0.2 M

K I

solution. The liberated

I_{2}

required

400

mL of

1.0 M

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

for complete reaction. The percentage purity of

C u S O_{4}

in the 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 :

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 amount of

I_{2}

liberated in the reaction of

10

ml of

0.02 M

solution with excess

K I

is :

Join BYJU'S Learning Program

I2 obtained from 0.1 mol of CuSO4 required 100 mL of 1M hypo solution. Hence, mole percentage of pure CuSO4 is:

The correct option is A 100The reactions that occur are given below:2CuSO4+4KI→Cu2I2+I2+2K2SO4 I2+2Na2S2O3→2NaI+Na2S4O6 2CuSO4≡2Na2S2O3 CuSO4≡Na2S2O3 100 mL of 1 M hypo =0.1 mole hypo ≡0.1 mol pure CuSO4 Hence, it is 100% pure.

$I_{2}$ obtained from $0.1$ mol of $C u S O_{4}$ required $100$ mL of $1$ M hypo solution. Hence, mole percentage of pure $C u S O_{4}$ is: