The molality and molarity of a solution of a glucose in water which is labelled as $10 \%$ $(w / w)$ are…
The molality and molarity of a solution of a glucose in water which is labelled as $10 \%$ $(w / w)$ are respectively (density of solution $\left.=1.2 \mathrm{~g} \mathrm{~mL}^{-1}\right)$
$0.57 \mathrm{~m}, 0.517 \mathrm{M}$
$0.67 \mathrm{~m}, 0.617 \mathrm{M}$
$0.617 \mathrm{~m}, 0.67 \mathrm{M}$
$0.517 \mathrm{~m} ; 0.57 \mathrm{M}$
Solution
Mass by mass percentage is the mass of solute dissolved in $100 \mathrm{~g}$ of the solution.
Mass of solution $=100 \mathrm{~g}$
Mass of glucose $=10 \mathrm{~g}$
Mass of solvent $=90 \mathrm{~g}$
Molar mass of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$
$=12 \times 6+1 \times 12+16 \times 6=180 \mathrm{~g} / \mathrm{mol}$
$\begin{gathered}\text { Molality }=\frac{\text { Mass of solute }}{\text { Molar mass of solute }} \times \frac{1000}{\text { Mass of solvent }} \\ \text { (in g) }\end{gathered}$
$=\frac{10}{180} \times \frac{1000}{90}=0.617 \mathrm{~m}$
Also, volume of solution $=\frac{\text { Mass of solution }}{\text { Density of solution }}$
$=\frac{100 \mathrm{~g}}{1.2 \mathrm{~g} / \mathrm{mL}}=83.33 \mathrm{~mL}$
$\therefore$ Molarity $=\frac{10}{180} \times \frac{1000}{83.33}=0.67 \mathrm{M}$
