MyQuestionIcon

1

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

Stoichiometric Calculations

2.5 g sample ...

Question

$2.5$ 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:

A

2

ml

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

B

20

ml

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

C

1

ml

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

D

10

ml

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

Open in App

Solution

The correct option is B $2$ ml
$100$ mL of $0.02 M C u S O_{4}$ solution corresponds to $100 \times 0.02 = 2$ mmol.
Out of this, $10$ mL (i.e $0.2$ mmol) are used for the titration.
$I_{2} + 2 N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6} + 2 I^{-}$
$2 C u S O_{4} \equiv 2 C u I_{2} \equiv I_{2} \equiv 2 N a_{2} S_{2} O_{3}$
The volume of hypo required is $\frac{0.2}{0.1} = 2$ ml.

Hence, the correct option is $(A)$ .

Suggest Corrections

thumbs-up

1

Similar questions

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

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

10

0.02 M

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

0.1 M

hypo solution for its complete reduction.The amount of

I_{2}

liberated in the reaction of

10

0.02 M

solution with excess

K I

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

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

10

0.02 M

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

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}

100

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

10

0.02 M

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

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}

K C l O_{3}

are dissolved in conc. HCl and the solution was boiled. Chlorine gas evolved in the reactions was in then passed through a solution of KI and liberated iodine was titrated with 100 ml of hypo. 12.3 ml of same hypo solution required 24.6 ml of 0.5 iodine for complete neutralization. Calculate % purity of

K C l O_{3}

K C l O_{3}

are dissolved in cone. HCl and the solution was boiled. Chlorine gas evolved in the reaction was then passed through a solution of KI and liberated iodine was titrated with 100 mL of hypo. 12.3 mL of same hypo solution required 24.6 mL of 0.5 N iodine for complete neutralization. Calculate % purity of

K C l O_{3}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Stoichiometric Calculations

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Stoichiometric Calculations

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon