Question

Calculate the molality of a solution of Ethanol $\left(C_2{H}_{5}OH\right)$ in water in which the mole fraction of ethanol is 0.040.

• A. 4.42 mol/kg
• B. 2.31 mol/kg
• C. 5.23 mol/ kg
• D. none of these

Answer 1

Answer: (B)

2.31 mol/kg

Morality (m) $\frac{Noofmolesofsolute}{Massofsolventinkg}$

$\therefore m=\frac{weightofsolute\left(w_A\right)}{Molecularweightofsolute\left(M_A\right)}\times \frac{1000}{weightofsolvent\left(W_B\right)}$

$m=\frac{W_A}{M_A}\times \frac{1000}{W_B}\times \frac{M_B}{M_B}=\frac{n_A}{n_B}\times \frac{1000}{M_B}$

$n_A=$ no of moles of solute $m=\frac{n_A}{n_B}\times \frac{1000}{M_B}\times \frac{(n_A+n_B)}{(n_A+n_B)}$

$n_B=$ no of moles of solvent

$\therefore m=\frac{x_A}{x_B}\times \frac{1000}{M_B}...\left(1\right)$ $x_A=$ mole fraction of solute

$x_B=$ mole fraction of solvent

Given :- $x_A=0.040$ Also $x_A+x_B=1\Rightarrow x_B=0.960$

Ethanol is solute 1 water is solvent $\therefore M_B=19$

Putting values of $x_A,x_B\&M_B$ in (1), we get

$m=\frac{0.040}{0.960}\times \frac{1000}{18}\Rightarrow m=2.31mol/kg$