Question

The mole fraction of water in a sulphuric acid solution is 0.85. Calculate the molality of the solution:

Answer 1

Mole fraction of water in solution $=$ 0.85 mole fraction $H_{2}S{O}_4$ in solution
$=1-0.85=0.15$.
If $n_1$ is the number of moles of water and $n_2$ is the number of moles $H_{2}S{O}_4$ in the solution, then
Mole fraction of $H_{2}S{O}_4=\frac{n_2}{n_1+n_2}=0.15$
Molality of $H_{2}S{O}_4$ solution means the number of moles of $H_{2}S{O}_4$ present in 1000 of $H_{2}O$. Thus, we have, $w_1=1000$
g or $n_1=\frac{1000}{18}=55.55,n_2=1$
$\frac{n_2}{55.55+n_2}=0.15$
$n_2=0.15n_2+8.3325$
or $n_2=9.8$
$\therefore Molality=9.8m$.
Alternatively, If $n_1$ and $n_2$ are the number of the moles of water and $H_{2}S{O}_4$ respectively, then
$x_{water}=\frac{n_1}{n_1+n_2}=0.85$ ......... (i)
$x_{H_{2}S{O}_4}=\frac{n_2}{n_1+n_2}$
$=1-0.85=0.15$ ............ (ii)
Dividing (ii) by ()i
$\frac{n_2}{n_1}=\frac{0.15}{0.85}$
Now, $n_1=\frac{1000}{18}=55.55$
$\therefore n_2=\frac{0.15}{0.85}\times 55.55=9.8moles$
Hence, molality of solution $=$ 9.8 m.