Question : 10 Mg of a solution containing 0.3 kg Na2CO3/kg solution is ...

The initial concentration of the solution = 0.3 kg/kg solution

or: c_{1} = 0.3/(1 – 0.3) = 0.428 kg/kg water.

The final concentration of the solution, c_{2} = (21.5/100) = 0.215 kg/kg water.

The feed of 10 Mg of solution contains (10 × 0.3) = 3 Mg of anhydrous salt and (10 – 3) = 7 Mg of water.

Thus: the initial mass of solvent in the liquid, w_{1} = (7 × 1000) = 7000 kg.
3 per cent of the original solution or (10 × 1000 × 3)/100 = 300 kg is evaporated.

Thus the mass of solvent evaporated/mass of solvent in initial solution is given by:

E = [300/(10 × 1000)] = 0.03 kg/kg solution.

The final mass of solvent in the liquid, w_{2} = (7000 – 300) = 6700 kg.

The molecular mass of Na_{2}CO_{3} = 106 kg/kmol and the molecular mass of Na_{2}CO_{3}.10H_{2}O = 286.2 kg/kmol and hence the molecular mass of hydrate/molecular mass of anhydrous salt is given by:

R = (286.2/106) = 2.7

Therefore from equation 15.22: y =Rw_{1}[c_{1}-c_{2}(1-E)]/[1-c_{2}(R-1)] and by substituting, the yield is:

y = (2.7 × 7000)[0.428 – 0.215(1 – 0.03)]/[1 – 0.215(2.7 – 1.0)]

=6536 kg .

equation 15.22: y=Rw_{1}\mathbf{\frac{c_{1}-c_{2}(1-E)}{1-c_{2}(R-1)}}