Answer:
The answer to your question is 0.22
Explanation:
Data
Acetonitrile (CH₃CN) density = 0.786 g/ml
Methanol (CH₃OH) density = 0.791 g/ml
Volume of CH₃OH = 22 ml
Volume of CH₃CN = 98.4 ml
Process
1.- Calculate the mass of Acetonitrile and the mass of Methanol
density = mass/ volume
mass = density x volume
Acetonitrile
mass = 0.786 x 98.4
= 77.34 g
Methanol
mass = 0.791 x 22
= 17.40 g
2.- Calculate the moles of the reactants
Acetonitrile molar mass = (12 x 2) + (14 x 1) + (3 x 1)
= 24 + 14 + 3
= 41 g
Methanol molar mass = (12 x 1) + (4 x 1) + (16 x 1)
= 12 + 4 + 16
= 32 g
Moles of Acetonitrile
41 g ----------------- 1 mol
77.34g ------------ x
x = (77.34 x 1) / 41
x = 1.89 moles
Moles of Methanol
32 g -------------- 1 mol
17.40 g --------- x
x = (17.40 x 1)/32
x = 0.54 moles
3.- Calculate the mole fraction of Methanol
Total number of moles = 1.89 + 0.54
= 2.43
Mole fraction = moles of Methanol / total number of moles
Mole fraction = 0.54/ 2.43
Mole fraction = 0.22
