Now
1.45 = [mass in octanol / 100]/ [mass in water/ 50]
mass in octanol / mass in water = 2.9
Mass in octanol = 2.9 × mass in water
Total mass = 3.5 g
3.9× mass in water = 3.5 g
Mass in water = 3.5/ 3.9
= 0.9 g
Molarity of the aqueous solution
= (0.9/ 101.19)/ 0.05
= 0.177 M
So this is the base concentration
Here 50 ml of 0.177M of water solution present so to neutralize volume of HCl required
= (0.177 × 50) / 0.5
=17.7 ml
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