Question

The Boiling Point Diagram of Methanol-Water (Figure 5.6) shows Mole Fraction of Water versus Temperature, °C. Using the following atomic weights and the density of methanol of 0.792 g/ml, a 41.5 % v/v methanol in water solution (41.5 ml methanol and 58.5 ml water) is what mole fraction of water? Round to 2 decimal points with no units. Remember that the density of water is defined in the metric system as 1. (C = 12, H = 1, N = 14, O = 16)

Answer 1

Solution :-

We have 41.5 % v/v% of mixture of methanol and water

Volume of methanol = 41.5 ml

Volume of water = 58.5 ml

Density of methanol =0.792 g/ml

Density of water = 1.0 g/ml

Molar mass of methanol CH3OH = 32 g/mol

Molar mass of water = 18 g/mol

Mole fraction of water = ?

Lets calculate the mass of each

Mass = volume x density

Mass of methanol = 41.5 ml * 0.792 g /ml = 32.87 g

Mass of water = 58.5 ml *1.0 g/ml = 58.5 g

To find the mole fraction of water we first need to calculate the moles of methanol and water

Moles = mass / molar mass

Moles of methanol = 32.87 g / 32 g per mol

                                  = 1.03 mol

Moles of water = 58.5 g / 18 g per mol

                             = 3.25 mol

Mole fraction of water = moles of water /[ moles of water + moles of methanol]

                                         = 3.25 mol / [3.25 mol + 1.03 mol ]

                                         = 0.76

Therefore the mole fraction of water is   0.76