Question

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 1

Mole fraction of C₂H₅OH = \(\frac {\text {Number\  of \ moles\  of}\ C_2H_5OH}{\text {Number\  of \ Moles\  of\  Solution}}\)
\(0.040 = \frac {n_{C_2H_5OH} }{ n_{C_2H_5OH} + n_{H_2O}}  \)     ......(1)
Number of moles present in 1 L water:
\(n_{H_2O} = \frac {1000\ g }{8 \ g\  mol^{-1}}\)
\(n_{H_2O} = 55.55\  mol\)
Substituting the value of nH₂O in equation (1),
\(\frac {n_{C_2H_5OH}}{{ n_{C_2H_5OH}}} + 55.55 = 0.040\)
\(n_{C_2H_5OH} = 0.04n_{C_2H_5OH} + (0.040).(55.55)\)
\(0.96n_{C_2H_5OH} = 2.222\ mol\)
\(n_{C_2H_5OH} = \frac {2.222 }{ 0.96} \ mol\)
\(n_{C_2H_5OH} = 2.314 \ mol\)
∴ Molarity of solution = \(\frac {2.314\  mol }{1 \ L}\) = \(2.314\  M\)