Question

A solution is prepared by adding 60 g of methyl alcohol to 120 g of water. Calculate the mole fraction of methanol and water.

• A. 0.2,0.8
• B. 0.22,0.78
• C. 0.25,0.75
• D. 0.5,0.5

Answer 1

Answer: (B)

0.22,0.78

Given conditions ⇒

Mass of the Methanol = 60 g.

Molar mass of the Methanol = 32 g/mole.

∵ No. of moles of Methanol = $\frac{Mass}{\text{Molar Mass}}$

∴ No. of moles = 60/32

= 1.875 mole.

Mass of the Water = 120 g.

Molar mass of the water = 18 g/mole.

∴ No. of moles of Water = $\frac{Mass}{\text{Molar mass}}$.

= 120/18

= 6.67 mole.

Now, Total number of moles in a solution = No. of moles of Methanol + No. of moles of water.

= 1.875 + 6.67

= 8.545 moles.

Now,

Mole Fraction of Methanol in the solution = $\frac{\text{No. of moles of Methanol}}{\text{Total number of moles in a solutions}}$.

= 1.875/8.545

= 0.22

Mole Fraction of Water in the solution = $\frac{\text{No. of moles of water}}{\text{Total number of moles in a solution}}$.

= 6.67/8.545

= 0.78