Question

4. A mixture contains 9.2 g of ethanol (CH3OH) and 18 g of water (H2O). Thus, mole fraction of ethanol
in the mixture is
(A) 0.2
(B) 0.1
(C) 0.25
(D) 0.167​

Answer 1

Answer:

D

Explanation:

Answer 2

Mass of ethanol = 9.2g

Moles of ethanol = Given mass / Molar mass = 9.2/46 = 0.2 mol

Mass of H2O = 18g

Moles of H2O = Given mass / Molar mass = 18/18 = 1 mol

Mole fraction of ethanol = No. of moles of ethanol/Total no. of moles present in soution = 0.2/1.2 = 0.167

Option (D) is correct.