Question

1, Lemonade contains 20% sucrose, 2% salt (NaCl), and 0.5% citric acid (w/v each) in water. Calculate the mole fraction for each of these components (sucrose, salt, citric acid and water).

2. 1000 kg of orange juice containing 15% solids is used to prepare orange juice concentrate in two stages. In the first stage the orange juice is passed through evaporator to remove water. The product from evaporator contains 45% solids. This partially concentrated juice is then fed into a second stage mixer where sugar is added to the juice in such a way that final solid content of the concentrate is 55%. Calculate the amount of a) water removed, b) sugar added c) final product.

Answer 1

1. Let us assume the whole quantity of lemonade is 100g

therefore it has 20g sucrose, 2g of NaCl and 0.5g of citric acid rest is water which is equivalent to 100-22.5 = 77.5g

No. of moles of each constituents:

Sucrose , C12H22O11 mol wt: 342.295g/mol , NaCl = 58.44g/mol, Citric acid, C6H8O7 = 192.124 g/mol, water = 18g/mol

No. of mole of sucrose : 20g/ 342.295g/mol = 0.0584 mol

No. of mole of NaCl = 2/58.44g/mol = 0.0342 mol

no. of moles of citric acid = 0.5 / 192.124 = 0.0026 mol

No. of mol of water = 77.5 /18 = 4.3055 mol

Total No. of moles (mles of sucrose +citric acid + water + NaCl) = 4.4007 mol

Mole fraction = moles of A/ total no. of moles

Mole fraction of sucrose = 0.0584/4.4007 =0.0132 mol

mole fraction of NaCl = 0.0342/4.4007 = 0.0077 mol

mole fraction of citric acid= 0.0026/4.4007 = 0.005 mol

mol fraction of water = 4.3055/4,4007 = 0.9784 mol

2. lets prepare a flow chart

15% solid ------> evaoprate to give 45% solid --------> sugar added to give 55% solids

total 1000kg orange juice as rest 15 % is solid which means 150 kg solid and 850 kg water from this evaporation 45% solid remains now consider 45 g solid in 100g mixture therfore 55 g of water is present

which means 300kg of water dissolve to make 45-15 = 30 g

850-300 = 550kg of water evaporate.

sugar added