Question

An aqueous sugar syrup of mass 224.2 g contains 34.2 g of sugar (C12 H22 O11) calculate
1. The molality of solution
2. The mole fraction of solute

Answer 1

Mass of water = 224.2 – 34.2 = 190g


Assuming density of water = 1g/ml


Volume of solution = Mass/Density =  190/1 = 190ml

Volume in litres = 190/1000 = 0.19 L



Molar mass of C12 H22 O11 = 12x12 + 22 + 16x11 = 342



Moles of C12 H22 O11 = 34.2/342 = 0.1 moles



1) Molarity = Number of moles/Volume in litres

  Molarity = 0.1/0.19 = 0.53M



2)  Molar mass of water (H2O) = 18g/mol


Moles of H2O = 190/18 = 10.56



Total number of moles = moles of C12 H22 O11 + moles of H2O


                                    = 0.1 + 10.56 = 10.66




The mole fraction of solute = 0.1/10.66 = 0.01

Answer 2

1. Molarity of given solution = 0.53 M