Question

$K_f$ for water is $1.86\, K \,kg mol^{-1}$. If your automobile radiator holds $1.0\, kg$ of water, how may grams of ethylene glycol $(C_2H_6O_2)$ must you add to get the freezing point of the solution lowered to $-2.8^{\circ}C$ ?

• A. $72\, g$
• B. $93\, g$
• C. $39\, g$
• D. $27\, g$

Answer 1

Answer: (B)

$\Delta T_{f} = i × k_{f }× m$
$2.8 = 1 × 1.86 × \frac{x}{62 \times1}$
$x = \frac{2.8 \times 62}{1.86} = 93 \,gm$