Question

A solution is made by dissolving 19.5 grams of glucose (C6H12O6) in .245 kilograms of water. If the molal freezing point constant for water (Kf) is -1.86 °C/m, what is the resulting Δtb of the solution? Please show work so I can understand it better :)

Answer 1

First you must calculate the molal = \[\frac{ 19.5 }{ 180 }\times \frac{ 1000 }{ 245 }=0.442\]
Then you calculate \[\Delta Tf= molal \times Kf\]