MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Ostwald Dilution Law

1 g of a samp...

Question

$1$ g of a sample of $N a O H$ was dissolved in $50$ mL of $0.33 M$ alkaline solution of $K M n O_{4}$ and refluxed till all the cyanide was converted into $O C N^{⊝}$ . The reaction mixture was collected and its $5$ mL portion was acidified by adding $H_{2} S O_{4}$ in excess and then titrated to endpoint against $19.0$ mL of $0.1 M$ $F e S O_{4}$ solution. The percentage purity of $N a C N$ sample is:

A

55.95 %

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

B

65.95 %

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

C

75.95 %

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

D

85.95 %

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

Open in App

Solution

The correct option is C $75.95 %$
In the alkaline medium, the reaction is
$\Rightarrow M n O_{4}^{⊝} + 3 e^{-} \to M n O_{2}$ ( $∵ n$ factor $= 3$ )
$\Rightarrow$ mEq of $N a C N = (0.33 \times 3 \times 50) - (0.1 \times 1 \times 19) \times \frac{50}{5} = 31$
Also, $C N^{⊝} + O_{2} \to O C N^{⊝} + 2 H^{⨁} + 2 e^{-}$
$\Rightarrow$ mmol of $N a C N = \frac{31}{2} = \frac{W e i g h t}{49} \times 1000$
$\Rightarrow$ Weight $= \frac{31}{2} \times \frac{49}{1000} = 75.95 %$ = $N a C N$ purity.

Suggest Corrections

thumbs-up

2

Similar questions

2.0

N a C N

is dissolved in

50

0.3 M

K M n O_{4}

and heated strongly to convert all the

C N^{⊝}

O C N^{⊝}

. The solution after acidification with

H_{2} S O_{4}

500

0.05 M

F e S O_{4}

. The percentage purity of

N a C N

in the sample is :

Q. 50 g of sample of

C a (O H)_{2}

is dissolved in 50 ml of 0.5 N HCl solution. The excess HCl was titrated with 0.25 N NaOH solution. The volume of NaOH used was 20 ml. What is % purity of

C a (O H)_{2}

Q. An impure sample of

P_{2} O_{3}

1.5

g was dissolved in water and warmed till whole of

P_{2} O_{3}

undergoes disproportionation in to

P H_{3}

H_{3} P O_{4}

. The solution was then boiled to remove

P H_{3}

from it and finally cooled to room temperature and made

100

10

mL of this solution was mixed with

1 O

0.6

N a O H

\frac{1}{3}

of this solution required

3.6

0.05

H_{2} S O_{4}

for back titration. What is

%

P_{2} O_{3}

50

g of a sample of

C a {(O H)}_{2}

50

0.5 N

H C l

solution. The excess of

H C l

was titrated with

0.3 N N a O H

. The volume of

N a O H

20

cc. The percentage purity of

C a {(O H)}_{2}

1 M N a O H

solution was slowly added into

1000

183.75

H_{2} S O_{4}

solution and the following plot was obtained. The percentage purity of

H_{2} S O_{4}

sample and slope of the curve respectively, are

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Oswalds Dilution Law

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Ostwald Dilution Law

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon