Question

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 1

Molar mass of ethylene glycol $[C_2H_4(OH)_2] = 2 \times12 + 6 \times 1 + 2\times16$
$= 62\, gmol ^{- 1}$

Number of moles of ethylene glycol $=\frac{222.6G}{62gmol^{-1}}$
$=3.59 mol$

Therefore, molality of the solution $=\frac{3.59mol}{0.200kg}$
$=17.95 m$

Total mass of the solution $= (222.6 + 200) g$
$= 422.6 g$
Given,
Density of the solution $= 1.072 g mL^{ - 1}$

∴Volume of the solution $= \frac{422.6 g}{1.072gmL^{-1}}$

$=3942.22mL$

$=0.3942\times10^{-3}L$

⇒Molarity of the solution $= \frac{3.59 mol}{(0.3942\times10^{-3}L)}$
$=9.11 M$