## anonymous 4 years ago What is the freezing point of an aqueous solution that boils at 103.7 degrees celsius?

In this case, you need to take your equations for boiling point elevation and freezing point depression and combine them...$\Delta T_b=K_b*m*i$$\Delta T_f=K_f*m*i$Solve for (m * i) in the first equation, and then substitute that into the second to get...$\Delta T_f=K_f*\frac{\Delta T_b}{K_b}$K_f for water is 1.86 K/m, and K_b for water is 0.512K/m. $\Delta T_f=(1.86 K*C^{-1})*\frac{103.7C-100C}{0.512K*C^{-1}}$If we solve for deltaT_f, we get 13.4C, and since the normal freezing point for water is 0C, the depressed freezing point is...$-13.4 C$