Question

$3.0g$ of oxalic acid$[(C{O}_{2}H{)}_2\cdot 2H_{2}0]$ is dissolved in a solvent to prepare a $250mL$ solution. The density of the solution is $1.9g/mL$. The molality and normality of the solution, respectively, are closest to

• A. $0.10molk{g}^{-1}$ and $0.38N$
• B. $0.10molk{g}^{-1}$ and $0.19N$
• C. $0.05molk{g}^{-1}$ and $0.19N$
• D. $0.05molk{g}^{-1}$ and $0.09N$

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
$=\underset{\downarrow }{\text{mass of solution}}-\underset{\downarrow }{\text{ Mass of solute}}$
Volume $\times$ density $3g$
$=250mL\times 1.9g/mL=475g$
$\therefore$ Mass of solvent $=472g$
Molar mass of $(C{O}_{2}H{)}_2.2H_{2}O=126gmo{l}^1$
$\therefore$ Molality $=\frac{3\times 1000}{472\times 126}$
$=0.05$mol $k{g}^{-1}$
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
$=\frac{\text{Molar mass}}{2}$
$=63g/$equi.
$\therefore$ Normality=$\frac{3\times 1000}{63\times 250}$
$=0.19$ (equivalents) $/l$
or $0.19N$