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...
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