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 \(= \frac{3.59 mol}{(0.3942\times10^{-3}L)}\)
\(=9.11 M\)

Answer 2

Step 1: Calculate the moles of each component.
Ethylene glycol (solute):
The molar mass of ethylene glycol \((C_2H_6O_2)\) = 62.07 g/mol
Moles of ethylene glycol = \(\frac{\text{mass}}{\text{molar mass}} =\frac{222.6 g}{62.07 g/mol} = 3.59 mol\)

Water (solvent):
The molar mass of water (\(H_2O\)) = 18.02 g/mol
Moles of water = \(\frac{\text{mass}}{\text{molar mass}} = \frac{200 g}{18.02 g/mol} = 11.11 \text{ mol}\)

Mass of solvent (water) = \(200 g\times\frac{1 kg}{1000 g}\) = 0.200 kg

Step 3: Calculate the molality.
Molality (m) = \(\frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}\)
Molality (m) = \(\frac{3.59\ mol}{0.200\ kg}\) = 17.95 mol/kg

Step 4: Find the total volume of the solution.
Total mass of solution = mass of ethylene glycol + mass of water = 222.6 g + 200 g = 422.6 g
Density (ρ) = 1.072 g/mL
Volume (V) = \(\frac{\text{mass}}{\text{density}}\)
Volume (V) = \(\frac{422.6 g}{1.072 g/mL}\) = 394.2 mL (convert mL to L by dividing by 1000)
Volume (V) = 0.3942 L

Step 5: Calculate the molarity.
Molarity (M) = \(\frac{\text{moles of solute}}{\text{volume of solution (in L)}}\)
Molarity (M) = \(\frac{3.59\ mol}{0.3942\ L}\) = 9.11 M

So, the answer is 9.11 M.