Question

The weight of ethyl alcohol which must be added to 1.0 L of water so that the solution will freeze at $14^\circ F$ is ( $K_{f}$ of water = 1.86 K kg $mol^{- 1}$ )

• A. 263.11 g
• B. 247.31 g
• C. 236.11 g
• D. 281.01 g

Answer 1

Answer: (B)

$^{\text{o}} \text{C} = \frac{\left(^{\text{o}} \text{F} - 32\right) \times 5}{9}$

$14^\circ F=-10^\circ C$

So $\Delta T=10$

$\Delta T=K_{f}\times \frac{w_{B} \times 1000}{M_{B} \times w_{A}}$

$10=1.86\times \frac{w_{B} \times 1000}{46 \times 1000}$

$w_{B}=247.31$ g.