MyQuestionIcon

1

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

Equivalent Mass

Mg can reduce...

Question

$M g$ can reduce $N O_{3}^{-}$ to $N H_{3}$ in basic medium.
$N O_{3}^{-} (a q) + M g (s) + H_{2} O \to M g (O H)_{2} (s) + O H^{-} (a q) + N H_{3} (g)$
A $25.0 m L$ sample of $N O_{3}^{-}$ solution was treated with $M g (s)$ . The $N H_{3} (g)$ was passed into $100 m L$ of $0.15 N H C l$ .
The excess of $H C l$ required $32.10 m L$ of $0.10 N N a O H$ for neutralization. What was the molarity of $N O_{3}^{-}$ ions in the original sample?

A

0.47125

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

B

0.3025

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

C

3.025

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

D

0.5005

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

Open in App

Solution

The correct option is A $0.47125$
$Milliequivalents of N H_{3} formed = Milliequivalents of H C l used = 100 \times 0.15 - 32.10 \times 0.10 = 11.79$

Here, n-factor of $N H_{3}$ is $1$ (acid - base reaction).

For redox change,

$∴ Milliequivalents of N H_{3} for n-factor 8 = 8 \times 11.79 = 94.32$
$∴ Normality of N O_{3}^{-} = \frac{94.32}{25} = 3.77$
$Molarity of N O_{3}^{-} = \frac{3.77}{8} = 0.47125$

Suggest Corrections

thumbs-up

0

Similar questions

M g

N O_{3}^{-}

N H_{3}

in basic medium.

N O_{3}^{-} (a q) + M g (s) + H_{2} O \to M g (O H)_{2} (s) + O H^{-} (a q) + N H_{3} (g)

25.0 m L

N O_{3}^{-}

solution was treated with

M g (s)

N H_{3} (g)

was passed into

100 m L

0.15 N H C l

.
The excess of

H C l

32.10 m L

0.10 N N a O H

for neutralization. What was the molarity of

N O_{3}^{-}

ions in the original sample?

Q. In the reaction given,

M g (s) + N O_{3}^{⊝} + H_{2} O \to M g (O H)_{2} (s) + ⊝ O H (a q) + N H_{3} (g)

20

mL of sample of

N O_{3}^{⊝}

solution is treated with

M g

N H_{3} (g)

was passed into

50

0.1 M

H C l

H C l

30

0.1 M

K O H

for its neutralization, then the molarity of

N O_{3}^{⊝}

ions in the original sample will be :

Q. 2.0 g of chalk was dissolved in 10 ml of 4N HCl and the solution was made up to 100 ml. 25 ml of this solution required 18.75 ml of 0.2 N NaOH solution for complete neutralization. Calculate the percentage of pure

C a C O_{3}

in the sample of chalk?

1.5

g of chalk was treated with

10

4 N

H C l

. The chalk was dissolved and the solution was made to

100

25

mL of this solution required

18.75

0.2 N

N a O H

solution for complete neutralisation. Calculate the percentage of pure

C a C O_{3}

in the sample of chalk?

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}

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Equivalent Mass

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon