Question

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem

Answer 1

I honestly don't know where to start

Answer 2

first calculate moles of glucose
\[25.5g \times \frac{ 1mol }{ 180.1g }=0.1416 \]
Molality of the solution \[\frac{ 0.1416mol }{ 0.398 kg H _{2} O}=0.356molal\]
change in freezing point of solution =\[kfm=-1.86Cm^{-1} \times 0.356=-0.66C^o \]