Answer to Question #214315 in General Chemistry for Sheik Sharmila

Question

Find out the volume of 0.03 M HCl required to neutralize 150 ml of 0.02 M Ca(OH)2 solution.

Answer 1

Molar Mass of HCl = 36.458

= 0.03 × 36.458 = 1.094g

Molar Mass of 74.093g

= 0.02 × 74.093 = 1.482g

= 150 × 1.482g

= 222g

Ratio = 3:2

Hence = 222× 3/2

= 333g-1.094

= 331.91g