Question

The molar freezing point constant for water is $1.86^{o}Cmole^{- 1}$. If 342 gm of canesugar $(C_{12}H_{22}O_{11})$ are dissolved in 1000 $gm$ of water, the solution will freeze at.

• A. $- 1.86^{o}C$
• B. $1.86^{o}C$
• C. $- 3.92^{o}C$
• D. $2.42^{o}C$

Answer 1

Answer: (A)

$- 1.86^{o}C$

Answer 2

$\Delta T_{f} = 1.86 \times \left( \frac{342}{342} \right) = 1.86^{o}$; ∴ $T_{f} = - 1.86^{o}C$.