Question

question 3
3 points
automotive antifreeze contains ethylene glycol, ch₂(oh)ch₂(oh), a nonvolatile nonelectrolyte, in water. calculate the freezing point of a 25.0% by mass solution of ethylene glycol in water. ( k_f = 1.86 , ^circ\text{c/m} ) and ( k_b = 0.51 , ^circ\text{c/m} ).
( circ , 2.7 , ^circ\text{c} )
( circ , -2.7 , ^circ\text{c} )
( circ , -10.0 , ^circ\text{c} )
( circ , 10.0 , ^circ\text{c} )

Explanation:

Step1: Define mass of solution

Assume mass of solution is 100g. Then mass of ethylene glycol (solute) is 25.0g, mass of water (solvent) is \(100 - 25 = 75\)g \(= 0.075\)kg.

Step2: Calculate moles of ethylene glycol

Molar mass of \( \text{CH}_2(\text{OH})\text{CH}_2(\text{OH}) \) (ethylene glycol) is \( 2\times12.01 + 6\times1.008 + 2\times16.00 = 62.07\)g/mol. Moles \( n=\frac{25.0\mathrm{g}}{62.07\mathrm{g/mol}}\approx0.4028\)mol.

Step3: Calculate molality (m)

Molality \( m=\frac{\text{moles of solute}}{\text{kg of solvent}}=\frac{0.4028\mathrm{mol}}{0.075\mathrm{kg}}\approx5.37\)m.

Step4: Calculate freezing point depression (\(\Delta T_f\))

For non - electrolyte, \( i = 1 \). \(\Delta T_f=K_f\times m\times i\). Given \( K_f = 1.86^{\circ}\text{C/m}\), \( m\approx5.37\)m, \( i = 1 \). So \(\Delta T_f=1.86\times5.37\times1\approx10.0^{\circ}\text{C}\).

Step5: Calculate freezing point of solution

Freezing point of pure water is \( 0^{\circ}\text{C} \). Freezing point of solution \( T_f = 0-\Delta T_f=- 10.0^{\circ}\text{C}\).

Answer:

\(-10.0^{\circ}\text{C}\) (corresponding to the option: \(-10.0^{\circ}\text{C}\))