CameraIcon

SearchIcon

MyQuestionIcon

1

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

Analgesics

1.725 g of a ...

Question

1.725 g of a metal carbonate is mixed with 300 ml of N/10 HCl. 10 ml of N/2 sodium hydroxide were required to neutralise excess of acid. Calculate the equivalent mass of metal carbonate.

Open in App

Solution

Suggest Corrections

thumbs-up

3

Similar questions

1.725 g

of a metal carbonate is mixed with

300 m L

\frac{N}{10} H C l . 10 m L

\frac{N}{2}

sodium hydroxide were required to neutralise excess of the acid. Calculate the equivalent mass of the metal carbonate.

3.45 g

of a metallic carbonate were mixed with

240 m L

N / 4 H C l

. The excess acid was neutralised by

50 m L

N / 5 K O H

solution. Calculate the equivalent mass of the metal.

1.0 g

carbonate of a metal was dissolved in

50 m L N / 2 H C l

solution. The resulting liquid required

25 m L

N / 5 N a O H

solution to neutralise it completely. Calculate the equivalent mass of the metal carbonate.

2 g

of a metal carbonate were dissolved in

50 m L

N H C l

100 m L

0.1 N N a O H

were required to neutralise the resultant solution. Calculate the equivalent mass of the metal carbonate.
(b) How much water should be added to

75 m L

3 N H C l

to make it a normal solution?

2

g of metal carbonate is neutralised completely by

100

0.1 N

H C l

. The equivalent mass of metal carbonate is :

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Neurologically Active Drugs

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon