Question

Calculate the mole fraction and molality of $HN{O}_3$, in a solution containing 12.2% $HN{O}_3$.
[Given - atomic masses: $H=1,N=14,O=16$]

Answer 1

Let us consider that there is 100 grams of $HN{O}_3$ solution.

$\therefore$ Mass of $HN{O}_3+$ Mass of $H_{2}O=100g$

$\because$ Mass of $HN{O}_3=12.2g$

$\therefore 12.2g+$ Mass of $H_{2}O=100g$

$\therefore$ Mass of $H_{2}O=100-12.2=87.8g$

$n_{H_{2}O}=\frac{W_{H_{2}O}}{M_{H_{2}O}}=\frac{87.8}{18}=4.87mol$

Number of moles of $HN{O}_3$

$n_{HN{O}_3}=\frac{W_{HN{O}_3}}{M_{HN{O}_3}}=\frac{12.2}{63}=0.1936mol$

$\therefore$ Total moles in the solution,

$n_{H_{2}O}+n_{HN{O}_3}=4.87+0.1936=5.06mol$

$\therefore$ Mole fraction of $HN{O}_3=\frac{\text{No. of moles of}HN{O}_3}{\text{Total number of moles (in soln.)}}$

$=\frac{0.1936}{5.06}=0.0383$

Now, Molality of $HN{O}_3=\frac{\text{Moles of solute}\left(HN{O}_3\right)}{\text{Mass of solvent in gram}}\times 1000$

$=\frac{0.1936\times 1000}{87.8}=2.205molk{g}^{-1}$

Hence, mole fraction and molality of HNO${}_3$, in a solution containing 12.2% HNO${}_3$ is 0.0389 and 2.205 mol kg${}^{-1}$ respectively