Question

Ethylene glycol (C\(_2\)H\(_6\)O\(_2\)) is used as an antifreeze in cars. In a certain place, it is desired that water freezes at 258K. How much weight of ethylene glycol is needed to prevent separation of ice from 500 g of water? \[ (K_f { of water} = 1.86 { K kg mol}^{-1}) \] 
 

• A. 125 g
• B. 62.5 g
• C. 250 g
• D. 175 g

Hint:

For freezing point depression calculations, always remember the formula \( \Delta T_f = i K_f m \) and correctly determine the molality based on the solvent mass.

Answer 1

The Correct Option is C

The freezing point depression is given by:
\[ \Delta T_f = i K_f m \]
where:
- \( \Delta T_f \) = \( 273 - 258 = 15 \) K (change in freezing point)
- \( K_f \) = 1.86 K kg mol\(^{-1}\) (cryoscopic constant for water)
- \( m \) = molality of solute (moles of solute per kg of solvent)
- \( i \) = 1 (since ethylene glycol is a non-electrolyte)
First, we calculate the molality:
\[ m = \frac{\Delta T_f}{K_f} = \frac{15}{1.86} = 8.06 \text{ mol/kg} \]
Since we have 500 g (0.5 kg) of water:
\[ \text{Moles of ethylene glycol required} = 8.06 \times 0.5 = 4.03 \text{ moles} \]
The molar mass of ethylene glycol (C\(_2\)H\(_6\)O\(_2\)) is:
\[ (2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \text{ g/mol} \]
Thus, the required mass of ethylene glycol:
\[ 4.03 \times 62 = 250 \text{ g} \]