Question

$3.0\,g$ of oxalic acid$ [(CO_2H)_2 \cdot 2H_20]$ is dissolved in a solvent to prepare a $250\,mL$ solution. The density of the solution is $1.9\, g/mL$. The molality and normality of the solution, respectively, are closest to

• A. $0.10\ mol \,kg^{-1}$ and $0.38\,N$
• B. $0.10\ mol\, kg^{-1}$ and $0.19\,N$
• C. $0.05\ mol\, kg^{-1}$ and $0.19\,N$
• D. $0.05\ mol\, kg^{-1}$ and $0.09\,N$

Answer 1

Answer: (C)

Molality $ = \frac{\text{Mass of solute} \times 1000}{\text{Molar mass of solute} \times {\text{Mass of solvent}}}$
Mass of solvent
$= \underset{\ce{v}}{\text{mass of solution}} - \underset{\ce{v}}{\text{ Mass of solute}}$
Volume $\times$ density $\quad\,\,\,\,3\,g$
$= 250\,mL \times 1.9\,g/mL = 475\,g$
$\therefore $ Mass of solvent $= 472\,g$
Molar mass of $(CO_2H)_2.2H_2O= 126\,g \,mol^{1}$
$\therefore $ Molality $=\frac{3\times1000}{472 \times126}$
$=0.05\, $mol $kg^{-1} $
Normality
$ = \frac{\text{Number of equivalents of solute}}{\text{Volume of solution (l)}}$
$= \frac{\text{Mass of solute} (g ) \times 1000}{\text{Equivalent mass of solute} \times {\text{Volume of solution}(mL)}}$
Equivalent mass of oxalic acid
$= \frac{\text{Molar mass}} {2}$
$ = 63\, g/$equi.
$\therefore $ Normality=$\frac{3\times1000}{63\times250} $
$=0.19$ (equivalents) $/l$
or $0.19\,N$