Question

What is the freezing point of an aqueous solution that boils at 103.7 degrees celsius?

Answer 1

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...\[-13.4 C\]