MyQuestionIcon

1

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

Stoichiometric Calculations

5 g sample of...

Question

$5$ g sample of brass was dissolved in one litre dil. $H_{2} S O_{4}$ . $20$ mL of this solution were mixed with $K I$ , liberating $I_{2}$ and $C u^{+}$ and the $I_{2}$ required $20$ mL of $0.0327$ N hypo solution for complete titration. Calculate the percentage of $C u$ in the alloy.

A

41.5 %

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

B

20.5 %

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

C

14 %

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

D

None of these

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

Open in App

Solution

The correct option is A $41.5 %$
$2 C u^{2 +} + 2 I^{-} ⇌ 2 C u^{+} + I_{2}$
$2 N a_{2} S_{2} O_{3} + I_{2} ⇌ N a_{2} S_{4} O_{6} + 2 N a I$
Thus, $1 m o l C u^{2 +} = 1 m o l N a_{2} S_{2} O_{3}$
Hence, total number of moles of copper in 20 ml ofsample $= \frac{20 \times 0.0327}{1000} = 6.54 \times 10^{- 4}$ mol.
Hence, total number of moles of copper in 1000 ml of sample $= \frac{1000 \times 6.54 \times 10^{- 4}}{20} = 0.0327$ mol.
Hence mass of copper $= 63.5 \times 0.0327 = 2.076$ g
Percentage of copper in sample $= \frac{2.076}{5} \times 100 = 41.5$ %.

Suggest Corrections

thumbs-up

1

Similar questions

Q. 5 g sample of brass was dissolved in one-liter

d i l . H_{2} S O_{4}

. 20 mL of this solution was mixed with

K I,

I_{2}

C u^{+}

I_{2}

required 20 mL of 0.0327 N hypo solution for complete titration. Calculate the percentage of Cu in the alloy.

5 g

of a sample of brass were dissolved in

1 l i t r e

H_{2} S O_{4} . 20 m L

of this solution were mixed with

K I

and liberated iodine required

20 m L

0.0327 N

hypo solution for titration. Calculate the amount of copper in the alloy.

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

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}

1.25

g of a sample of bleaching powder was dissolved in

100

ml of water and

25

ml of which are titrated with

K I

solution. The iodine liberated required

12.5

\frac{N}{25}

hypo-solution in titration. The percentage of chlorine in the sample of bleaching powder is:

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