Question

calculate the molality of the solution of $C{}_{2}H{}_{5}OH$ in water in which the mole fraction of $C{}_{2}H{}_{5}OH$ is 0.050

Answer 1

• The molality is given by the equation as:

$Molality\left(m\right)=\frac{Numberofmolesofsolute}{MassofsolventinKg}\times 100$

$m=\frac{x_2\times 1000}{x_1\times M_1}$

$where{x}_2=molefractionofsolute=0.050$

$x_1=molefractionofsolvent=1-0.050=0.95$

$M_1=molarmassofsolvent\left(water\right)=18g$

$Thus,m=\frac{0.050\times 100}{0.95\times 18}=2.92molperkg$

Alternate method

$Molality=\frac{No.ofmolesofsolute{n}_{solute}}{W_{solvent}\left(inkg\right)}\times 100$

$=\frac{n_{solute}}{W{\left(ingms\right)}_{solvent}}\times 1000$

$=\frac{n_{solute}}{\frac{W_{solvent}}{M_{solvent}}\times M_{solvent}}\times 1000$

$=\frac{n_{solute}}{n_{solvent}}\times \frac{1000}{M_{solvent}}$

The number of moles is directly proportional to its mole fraction, hence$n_{solvent}$ can be changed to $X_{solvent}$ also in the case of solute.

$Molality=\frac{X_{solute}}{X_{solvent}}\times \frac{1000}{M_{solvent}}$

$=\frac{X_{solute}}{1-X_{solute}}\times \frac{1000}{M_{solvent}}$

$=\frac{0.05}{1-0.05}\times \frac{1000}{18}$

$=2.924molal$

Hence the molality of the solution of $C_2{H}_{5}OH$ in water in with a mole fraction of $C_2{H}_{5}OH$ , is 0.050 is $2.924molal$ ​