133. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^–1, then what shall be the molarity of the solution?

Answer:

Consider ethylene glycol as solute and water as a solvent.
Weight of solute, W_B = 222.6 g
Molar mass, M_B = 24 + 6 + 32 = 62
W_B = 200 g = 0.200 kg.

Moles of solute,       ${\mathrm{n}}_{\mathrm{B}} = \frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}} = \frac{222.6}{62} = 3.59$

Molaity of ethylene glycol in H₂O

$$\begin{gathered} {\mathrm{m}}_{\mathrm{b}} =\frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Kg}. \mathrm{of} \mathrm{solvent}} \\ = \frac{3.59}{{\mathrm{W}}_{\mathrm{A}}} = \frac{3.59}{0.20} = 17.95 \mathrm{molal} \\ = 17.95 \mathrm{m} \end{gathered}$$

Total mass of solution
= 222.6 + 200 = 422.6 g
Density of solution
= 1.072 g/ ml Volume of solution

$$\begin{gathered} = \frac{\mathrm{Mass}}{\mathrm{Density}} = \frac{422.6}{1.072} = 394.2 \mathrm{mL} \\ = 0.3942 \mathrm{L} \end{gathered}$$

Molarity of solution,

$$\begin{gathered} {\mathrm{M}}_{\mathrm{b}} = \frac{\mathrm{Moles} \mathrm{of} \mathrm{solute}}{\mathrm{Vol}. \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{litres}} \\ = \frac{3.59}{0.3942} = 9.1\hspace{0.17em}\mathrm{M}. \end{gathered}$$