Question

The normality of a 500 ml Ca(OH)2 solution which contains 9.033×10ki power 22​

Answer 1

Answer: .6 equivalent /Litres

Explanation:

Normality is just another procedure to check concentration of solute in solution much similar like Molarity.

Now normality formula is

Normality = no of equivalent / volume in litres

Now,

Number of equivalent = given mass / equivalent mass

No of moles = given mass /molecular mass

For this we will consider Avogadro no of particles in a mole, and we've 9.033×10 to power 22​ particles of Calcium hydroxide.

Thus no of moles=

9.033* 10^22/ 6.022*10^ 23= 0.15 moles

Molar mass of Ca(OH)2 is 40+( 16+1)*2=74gms

So, 0.15 moles will be 0.15*74= 11.1 grams which will be given mass

Equivalent weight = molar mass / valence factor

= 74/2=37

For calcium hydroxide valence factor is 2.

No of equivalent = 11.1/ 37= 0.3

Given volume is 500ml or 0.5 litre

Thus normality = 0.3/ 0.5 = 0.6

Answer 2

The answer to above question is as shown below. I hope you understand....this is my first answer on Brainly app.....mark me the Brainliest......