Question

248 g of ethylene glycol (C\(_2\)H\(_6\)O\(_2\)) is added to 200 g of water to prepare antifreeze. What is the molality of the resultant solution?

• A. 5 m
• B. 10 m
• C. 20 m
• D. 40 m

Hint:

Molality = moles solute / kg solvent; unit is mol/kg.

Answer 1

The Correct Option is C

Step 1: Calculate moles of ethylene glycol
\[ Molar\, mass = 2 \times 12 + 6 \times 1 + 2 \times 16 = 62\, g/mol \] \[ moles = \frac{248}{62} = 4\, mol \] Step 2: Calculate mass of solvent in kg
\[ mass = 200\, g = 0.2\, kg \] Step 3: Calculate molality
\[ molality = \frac{moles\, solute}{kg\, solvent} = \frac{4}{0.2} = 20\, m \] Step 4: Conclusion
Molality of the solution is 20 m.