Question

A solution is made containing 48.7g of methanol (CH3OH) in 433.3g of water (H2O). What is the mole fraction of water? What is the mass percent of methanol? What is the molality of methanol?

Answer 1

Calculation of mole fraction of water

We know that, Mole fraction of water = n _H2O / n _H2O + n _CH3OH

Where, n _H2O is no. of moles of water and n _CH3OH is no. of moles of methanol.

Now, calculate no. of moles of water and methanol.

We have, no. of moles = Mass / Molar mass

No. of moles of water = 433.3 g / ( 18.02 g/mol ) = 24.046 mol

No. of moles of methanol = 48.7 g / ( 32.04 g/ mol ) = 1.520 mol

Mole fraction of water = n _H2O / n _H2O + n _CH3OH = 24.046 mol / ( 24.046 mol + 1.520 mol )

=  24.046 mol / 25.566 mol

= 0.9405

ANSWER : Mole fraction of water = 0.940

Calculation of mass percent of methanol

We have , % by mass = [ Mass of solute / Total mass of solution ] 100

mass percent of methanol = [ Mass of methanol / Total mass of solution ] 100

Total mass of solution = Mass of methanol + Mass of water = 48.7 g + 433.3 g = 482 g

Mass percent of methanol = [ 48.7 g / 482 g ] 100

= 10.1

ANSWER : Mass percent of methanol = 10.1 %

Calculation of molality of methanol solution

We have, molality of solution =  No. of moles of solute / Mass of solvent in kg

molality of methanol solution = No. of moles of methanol / Mass of water in kg

We have calculated,  No. of moles of methanol = 1.520 mol , Mass of water in kg = 0.4333 kg.

Substituting these values in above formula, we get

molality of methanol solution = 1.520 mol/ 0.4333 kg.

= 3.5079 mol / kg

ANSWER : molality of methanol solution = 3.51 mol / kg