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 = $\frac {2.314\ mol }{1 \ L}$ = $2.314\ M$