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