MyQuestionIcon

1

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

VSEPR Theory

5.5 g of a mi...

Question

$5.5 g$ of a mixture of $F e S O_{4} \cdot 7 H_{2} O$ and $F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O$ required $5.4 m L$ of $0.1 N K M n O_{4}$ solution for complete oxidation. Calculate the gram mole of hydrated ferric sulphate in the mixture.
(At. mass $H = 1, O = 16, S = 32, F e = 56$ ).

Open in App

Solution

Only $F e S O_{4} \cdot 7 H_{2} O$ will be oxidised by $K M n O_{4}$ .
Mol. mass of $F e S O_{4} \cdot 7 H_{2} O = 278$
As the conversion involves one electron,
$F e^{2 +} \to F e^{3 +} + e^{-}$
The eq. mass of $F e S O_{4} \cdot 7 H_{2} O = \frac{278}{1} = 278$
$5.4 m L$ of $0.1 N K M n O_{4}$
$\equiv 5.4 m L$ of $0.1 N F e S O_{4} \cdot 7 H_{2} O$ solution
Amount of $F e S O_{4} \cdot 7 H_{2} O = \frac{0.1 \times 278}{1000} \times 5.4 = 0.15 g$
Amount of $F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O = (5.5 - 0.15) = 5.35 g$
Mol. mass of $F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O = 562$
No. of $g$ moles of $F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O = \frac{M a s s}{M o l . m a s s}$
$= \frac{5.35}{562} = 0.00952$
$= 9.52 \times 10^{- 3}$ .

Suggest Corrections

thumbs-up

0

Similar questions

Q. 5.5 gm of a mixture of

F e S O_{4} . 7 H_{2} O

F e_{2} (S O_{4})_{3} .

9 H_{2} O

requires 5.4 ml of 0.1 N

K M n O_{4}

solution for complete oxidation. Calculate the number of gram moles of hydrated ferric sulphate in the mixture?

5.5

g of a mixture of

F e S O_{4} .7 H_{2} O

F e_{2} (S O_{4})_{3} .9 H_{2} O

5.4

0.1 N K M n O_{4}

solution for complete oxidation. Moles of hydrated ferric sulphate in mixture are

9.52 \times 10^{- x}

moles. The value of

x

5.5

g of a mixture of

F e S O_{4} .7 H_{2} O

F e_{2} (S O_{4})_{3} .9 H_{2} O

5.4

0.1

N

K M n O_{4}

solution for complete oxidation, then calculate the number of gram moles of hydrated ferric sulphate in mixture.
(ii) The vapour density of a mixture consisting of

N O_{2}

N_{2} O_{4}

38.3

{26.7}^{o} C

. Calculate the number of moles of

N O_{2}

100

g of the mixture.

0.108 g

of finely divided copper was treated with an excess of ferric sulphate solution until copper was completely dissolved. The solution after the addition of excess dilute sulphuric acid required

33.7

0.1 N K M n O_{4}

for complete oxidation. Find the equation which represents the reaction between metallic copper and ferric sulphate solution (Atomic mass of

C u = 63.6

F e = 56

F e S O_{4}

F e C_{2} O_{4}

was dissolved in dilute sulphuric acid. The complete oxidation of the reaction mixture required 40 mL of N/15

K M n O_{4} .

after the oxidation, the reaction mixture was reduced by Zn and

H_{2} S O_{4} .

On using the same solution of

K M n O_{4}

to oxidize the resulting solution, 25 mL was required. Calculate the ratio of equivalents of Fe in oxalate and Fe(II) sulphate. (Fe: 56)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon