Question

Sulfuric acid solutions containing 60 wt% H2SO4 (specific gravity =1.498) is to be diluted to a 4.00 molar solution (specific gravity=1.213) Calculate(a)the mass fraction of sulfuric acid in the product stream (b)taking 100 kg of feed solution, 60% sulfuric acid as a basis, calculate the kg of water for dilution and the kg of final acid

Answer 1

Basis: 100 gm of sulphuric acid   solution specific gravity = 1.498 = density of given substance/ density of water

Water density is taken as 1000 kg/m³

Density of solution = 1000*1.498= 1498 kg/m³ =1498 g/L

Volume of solution = 100/1498 L =0.067 L

Mass of sulphuric acid = 100*60/100 =60 gm

Molecular weight of sulphuric acid = 98 gm

Moles of sulphuric acid =Mass/Molecular weight= 60/98=0.6122

Molarity= moles of sulphuric acid/ L of solution = 0.6122/0.067=9.13M

Final molarity= 4M

This molarity is achieved by adding water to Sulfuric acid. This does not change the moles of sulphuric acid.

4= moles of Sulfuric acid/ Liter of solution

Volume of solution = 0.6122/4=0.15305L=153.05ml

Mass of final solution = volume* density = 153.05*1.213*1(g/cc)=185.6 gms

Mass of sulphuric acid = 60 gm

Mass fraction of sulphuric acid in 4M solution = 60/185.6=0.323

Water in the initial solution =100-60 =40 gm

Water in the final solution =185.6 gm

Water additionally added = 185.6-40 =145.6 gms

100gm of sulphuric acid solution require 149.6 gm of water for dilution

100 kg requires 149.8 kg of water for dilution

100gm of sulphuric acid solution contains 60 gm of sulphuric acid

100 kg of solution contains 60 kg of sulphuric acid