Question

The molar freezing point constant for water is $ 1.86{{\,}^{o}}\text{C/mol}\text{.} $ 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{{\,}^{o}}C $
• B. $ 1.86{{\,}^{o}}C $
• C. $ -3.92{{\,}^{o}}C $
• D. $ 2.42{{\,}^{o}}C $

Answer 1

Answer: (A)

Molality of cane sugar solution $ \frac{342}{342\times 1}=1\,m $ We know that, $ \Delta {{\Tau }_{f}}={{K}_{f}}.m=1.86\times 1={{1.86}^{o}} $ Hence, freezing point of solution $ =0.00-(1.86)=-1.86{{\,}^{o}}C $