Question

A solution is prepared by dissolving 15 ml of methanol in 150.0 ml water at 25 degree celsius. Assume the volumes are additive. The densities of methanol and water at this temperature are 0.782 g/ml and 1.00 g/ml, respectively. For the solute , calculete molarity, molality, and more fraction?

Answer 1

Solution :-

15 ml methanol with density 0.782 g /ml

150.0 ml water with density 1.00 g /ml

Therefore total volume of mixture is 150.0 ml + 15.0 ml = 165.0 ml = 0.165 L

Mass of methanol = volume * density

                                  = 15.0 ml * 0.782 g per ml

                                 = 11.73 g

Mass of water = 150.0 ml * 1.00 g /ml = 150.0 g = 0.150 kg

Now lets calculate the moles of methanol and water

Moles = mass / molar mass

Moles of methanol = 11.73 g / 32.04 g per mol = 0.3661 mol

Moles of water = 150.0 g / 18.0148 g per mol = 8.326 mol

Now lets calculate the molarity

Molarity = moles of solute / volume in liter

                = 0.3661 mol / 0.165 L

                = 2.22 M

Now lets calculate molality

Molality = moles of solute / kg solvent

                = 0.3661 mol / 0.150 kg

               = 2.44 m

Now lets calculate mole fraction of the methanol

Mole fraction = moles / total moles

                         = 0.3661 mol / (8.326 mol + 0.3661 mol )

                         = 0.0421