1 A mixture of alcohol and water is 2800 vv alcohol How many

Solution

1.

28% (v/v) means 28 L of solute (alcohol) is dissolved (or added) in 100 L of mixture.

Thus water in 100 L mixture = 100 - 28 = 72 L.

Thus 28 L alcohol needs 72 L water or 1 L alcohol needs 72/28 L water

Therefore 250 mL or 0.250 L of alcohol will need water = 0.250*72/28

                                                                                                    = 0.64 L of water

2.

% (m/m) salt = (mass of salt / mass of solution) * 100

                        = [28/ (28 + 60)]*100 = 32%

3.

33% (w/w) means 33 g of salt is dissolved (or added) in 100 g of mixture.

Thus weight of mixture containing 1 g salt is 100/33 g

Hence weight of mixture containing 300.0 g = 300.0*100/33 g = 909.1 g

4.

15% (w/w) water means 15 g of water is in 100 g of mixture.

Thus water present in 1 g mixture = 15/100 g

Hence water present in 38.6 g mixture = 38.6*15/100 g = 5.79 g