Question

What is the concentration of each of the following solutions? (a) The molality of a solution prepared by dissolving 25.0 g of H2SO4 in 1.30 L of water (b) The mole fraction of each component of a solution prepared by dissolving 2.25 g of nicotine, C10H14N2, in 80.0 g of CH2Cl2

Answer 1

a)

Molality of solute in a solution is,

Molality = number of moles of solute/ mass of solvent.

Remember it is mass of solvent, NOT the solution.

Molar weight of H2SO4 = 98 g per mol

Assuming H2SO4 used is 100% pure sample.

Number of moles of H2SO4 = 25 g / 98 g per mol

                                        = 0.255 mol

I assume that density of water is 1000 kg/m3 (1000 g per litre).

The mass of solvent present = 1.3 L x 1000 g/L

                                          = 1300 g.

                                          = 1.3 kg.

Therefore molality of the solution is,

Molality = 0.255 mols of H2SO4 / 1.3 kg of water

           = 0.1962 mol/kg

b) Molar weight of C10H14N2 = (12x10+1x14+14x2)

                                                = 162 g/mol

Molar weight of CH2Cl2   = (12x1+1x2+35.5x2)

                                         = 85 g/mol

Therefore number of C10H14N2 mols = 2.25 g / 162 g/mol

                                                      = 0.01389 mol

Number of CH2Cl2 mols = 80 g / 85 g/mol

                                   = 0.94118 mol

Therefore total mols peresent in solution

                               = 0.01389 mol + 0.94118 mol

                               = 0.95507 mol

Mole fraction of C10H14N2 = 0.01389 mol / 0.95507 mol

                                       = 0.0145

Therefore mole fraction of CH2Cl2 = 1 - 0.0145

                                                 = 0.9855