Q. The molar freezing point constant for water is $1.86{^\circ}C$/mol. 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) $1.86℃$
C) $- 3.92℃$
D) $2.42℃$
Explanation
(a)
Molality of cane sugar solution
$= \frac{342}{342 \times 1} = 1m$
We know that $\mathrm{\Delta}T_{f} = k_{f}.m$
=1.86 $\times$1
=1.86${^\circ}$
Hence , freezing point of solution
=0.00-(1.86)=-1.86$℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.