Q.The molal freezing point depression constant ( $K_f$ ) of water is $1.86 \text{ K kg mol}^{-1}$ . If $342 \text{ g}$ of cane sugar is dissolved in $1000 \text{ g}$ of water, the solution will freeze at:

-1.86^{\circ}C

1.86^{\circ}C

-3.42^{\circ}C

0^{\circ}C

The correct solution involves applying the fundamental concept to derive the final value step by step...

Already a member? Login here

Q.The molal freezing point depression constant (KfK_fKf​) of water is 1.86 K kg mol−11.86 \text{ K kg mol}^{-1}1.86 K kg mol−1. If 342 g342 \text{ g}342 g of cane sugar is dissolved in 1000 g1000 \text{ g}1000 g of water, the solution will freeze at: