Question

How many gram of CaCl2 are required to prepare 2L of it's semimolar solution.​

Answer 1

I think approximately 200 g

Answer 2

Answer:

here normality = 1/2

molar mass of CaCl2 = 40+71

= 111 gram

let mass of CaCl2 taken be x

molarity = x/111/2 = x/222M

normality = molarity * n-factor

1/2= x/222*2

1/2 = x/111

x= 111/2

Explanation: