Calculate the molarity of a solution of ethanol in water in which the

1.	Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
Answer» Mole fraction of C₂H₅OH = No. of moles of C₂H₅OH/No. of moles of solution n(C₂H₅OH) = n(C₂H₅OH)/n(C₂H₅OH)+n(H₂O) = 0.040 (Given) ……..Eqn. 1 We have to find the number of moles of ethanol in 1L of the solution but the solution is dilute. Therefore, water is approximately 1L. No. of moles in 1L of water = 1000g/18g mol^-1 = 55.55 mol Substituting n(H₂O) = 55.55 in eqn 1 n(C₂H₅OH)/n (C₂H₅OH) + 55.55 = 0.040 ⇒ 0.96n(C₂H₅OH) = 55.55 × 0.040⇒ n(C₂H₅OH) = 2.31 molHence, molarity of the solution = 2.31M

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Answer»

Mole fraction of C₂H₅OH = No. of moles of C₂H₅OH/No. of moles of solution

n(C₂H₅OH) = n(C₂H₅OH)/n(C₂H₅OH)+n(H₂O) = 0.040 (Given) ……..Eqn. 1

We have to find the number of moles of ethanol in 1L of the solution but the solution is dilute.

Therefore, water is approximately 1L.

No. of moles in 1L of water = 1000g/18g mol^-1 = 55.55 mol

Substituting n(H₂O) = 55.55 in eqn 1 n(C₂H₅OH)/n (C₂H₅OH) + 55.55 = 0.040

⇒ 0.96n(C₂H₅OH) = 55.55 × 0.040⇒ n(C₂H₅OH)

= 2.31 molHence, molarity of the solution = 2.31M