Q. The molar freezing point constant for water is $1.86℃\ mol^{- 1}$. If 342 g of cane sugar $(C_{12}H_{22}O_{11})$ is dissolved in 1000 g of water, the solution will freeze at
A) $- 1.86℃$
B) $- 2.86℃$
C) $+ 1.86℃$
D) $+ 2.86℃$
Explanation
(a)
$\mathrm{\Delta}T_{f} = 1.86 \times \frac{342}{342} = 1.86℃$
$\therefore T_{f} = T - \mathrm{\Delta}T_{f} = 0 - 1.86 = - 1.86℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.