Question

A solution is prepared by adding 60g of methyl alcohol to120 gm of water.calulate the mole fraction of methanol and water

Answer 1

Given conditions ⇒
Mass of the Methanol = 60 g.
Molar mass of the Methanol = 32 g/mole.

∵ No. of moles of Methanol = Mass/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 = Mass/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 = No. of moles of Methanol/Total number of moles in a solutions.
= 1.875/8.545
= 0.22

Mole Fraction of Water in the solution = No. of moles of water/Total number of moles in a solution.
 = 6.67/8.545
 = 0.78


Hope it helps.

Answer 2

Answer:

Explanation:

Given conditions ⇒

Mass of the Methanol = 60 g.

Molar mass of the Methanol = 32 g/mole.

∵ No. of moles of Methanol = Mass/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 = Mass/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 = No. of moles of Methanol/Total number of moles in a solutions.

= 1.875/8.545

= 0.22

Mole Fraction of Water in the solution = No. of moles of water/Total number of moles in a solution.

 = 6.67/8.545

 = 0.78

Hope it helps