Question

the density of 1.48 percent by weight Ca(OH)2 solution is 1.025 gram per ml find (a) the molarity of solution +b) volume of 0.1 N HCL required to neutralize 25 ml of Ca(OH)2 solution ​

Answer 1

Given:

d = 1.025 gm/l

wt % = 1.48 %

V2 = 25 ml

For HCl, N1 = 0.1

To Find:

(a) The molarity of Ca(OH)2 solution.

(b) The volume of given HCl to neutralize given volume of Ca(OH)2.

Calculation:

(a) Molarity = (10 × d × wt%)/ M.wt

⇒ M =  (10 × 1.025 × 1.48)/ 74

⇒ M = 0.205 M

(b) For Ca(OH)2, n = 2

- Normality of Ca(OH)2 = n × M

⇒ N2 = 2 × 0.205

⇒ N2 = 0.41 N

- Applying normality equation, we get:

N1V1 = N2V2

⇒ 0.1 × V1 = 0.41 × 25

⇒ V1 = 10.25/0.1

⇒ V1 = 102.5 ml

- So, the molarity of Ca(OH)2 solution is 0.205 M and the volume of HCl required to neutralize 25 ml of Ca(OH)2 is 102.5 ml