Question

Can someone please explain how to do this?
A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 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

omg soooo easy man

Answer 2

You need to use this formula:
\[m = \Delta T/K\]
where m = molality (moles of solute {glucose in this case} per kg of solvent {water in this case})
10.20 grams glucose / 180.16 g/mole = 0.05662 moles glucose
0.05662 moles glucose / 0.355 kg water = 0.1595 molality
where \[\Delta T\] = change in temperature (Celsius) = unknown
where K = freezing point constant = -1.86 Celsius

\[\Delta T = (0.1595)(-1.86) = -1.7005\]

The answer should be a freezing point depression (decrease) of 1.7 degrees Celsius