What are the concentrations of each of the following solutions?
a) The molality of a solution prepared by dissolving 25.0g of H2SO4 in 1.30L of water
b) The mole fraction of each component of a solution prepared by dissolving 2.25g of nicotine, C10H14N2, in 80.0 g of CH2Cl2
a) The molality is defined as follows,
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. So this can be tricky.
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
Therefore mole fraction of CH2Cl2 = 1 - 0.0145