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

Byju's Answer

Standard XII

Chemistry

Abnormal Colligative Properties

Calculate the...

Question

Calculate the number of millilitres of $N H_{3} (a q)$ solution $(d = 0.986$ g/ml $)$ containing $2.5 %$ by weight $N H_{3}$ , which will be required to precipitate iron as $F e (O H)_{3}$ in a $0.8 g$ sample that contains $50 % F e_{2} O_{3}$ .

0.344

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

3.44

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

17.24

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

10.34

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

Open in App

Solution

The correct option is D $10.34$ ml
The precipitation reaction is as follows:
$F e^{3 +} + 3 N H_{3} + 3 H_{2} O \to F e (O H)_{3} + 3 N H_{4}^{+}$

Moles of $F e_{2} O_{3}$ in sample $= \frac{0.80 \times 0.5}{160}$ $= 2.5 \times 10^{- 3}$

Moles of $F e^{3 +} = 2 \times 2.5 \times 10^{- 3} = 5 \times 10^{- 3}$

$M_{N H_{3}} = \frac{0.986 g / m l \times 1000 m l / l i t r e \times 0.025}{17}$
Therefore, $M_{N H_{3}} = 1.45$

$3 \times$ moles of $F e^{3 +} =$ moles of $N H_{3}$

Also, moles of $N H_{3}$ $= 1.45 \times V (i n L)$

$V = \frac{3 \times 5 \times 10^{- 3}}{1.45} = 10.34 \times 10^{- 3} L$ or $10.34 m l$

Suggest Corrections

Similar questions

Q. Calculate the number of millilitres of

N H_{3} (a q)

solution (d =

0.986

g/mL) containing

2.5 %

by weight

N H_{3}

, which will be required to precipitate iron as

F e (O H)_{3}

in a

0.8

g sample that contains

50 % F e_{2} O_{3}

(as nearest integer).

Q. A commercial sample (2.013 g) of NaOH containing

N a_{2} C O_{3}

as an impurity was dissolved to give 250 mL of solution. A 10 mL portion of the solution required 20 mL of 0.1 N

H_{2} S O_{4}

solution for complete neutralisation of

N a O H

. Calculate the percentage by weight of

N a_{2} C O_{3}

in the sample.[Write up to 2 decimal places]

1.20 g

sample of

N a_{2} C O_{3}

and

K_{2} C O_{3}

was dissolved in water to form

100 m L

of a solution.

20 m L

of this solution required

40 m L

0.1 N H C l

for complete neutralization. Calculate the weight of precipitate in the mixture if another

20 m L

of this solution is treated with excess of

B a C l_{2}

1.249

g of a sample of pure

B a C O_{3}

and impure

C a C O_{3}

containing some

C a O

was treated with dilute

H C l

and it evolved

168

mL of

C O_{2}

at NTP. From this solution,

B a C r O_{4}

was precipitated, filtered and washed. The dried precipitate was dissolved in dilute sulphuric acid and diluted to

100

mL.

10

mL of this solution when treated with

K I

solution liberated iodine, which required exactly

20

mL of

0.05 N

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

. Calculate the percentage of

C a O

in the sample.

Q. What volume (in mL) of

0.01 M A g N O_{3}

would be required to precipitate all the

I^{-}

200

mL of a solution that contained

24.32

ppt of

K I

Join BYJU'S Learning Program

Calculate the number of millilitres of NH3(aq) solution (d=0.986 g/ml) containing 2.5% by weight NH3, which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50%Fe2O3.

Calculate the number of millilitres of $N H_{3} (a q)$ solution $(d = 0.986$ g/ml $)$ containing $2.5 %$ by weight $N H_{3}$ , which will be required to precipitate iron as $F e (O H)_{3}$ in a $0.8 g$ sample that contains $50 % F e_{2} O_{3}$ .