Question

A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 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

@nincompoop

Answer 2

@abhisar

Answer 3

i = the van hoff factor means how many particles are being dissolved in your solvent.
our solute is glucose, and our solvent is water.

Answer 4

our van hoff factor is 1 because remember glucose doesn't form any ions in solution and doesnt break up, so it's just 1.

now we need to calculate the molality which is

\[molality= \frac{ moles of solute }{ kg solvent }\]

The most interesting thing is that they've given you the temperature change. when we dissolve a solute in a solvent it changes the freezing point and boiling point. this is called colliagitive properties which only depend on the amount of particles you put into your solvent.

we figure out the moles of glucose.

\[15.5 (g)_{glucose}*(\frac{ mol }{ 180_{molar mass glucose} }) = moles_{glucose}\]

245 grams of water. that's our solvent what we're dissolving glucose in. so what we do now is convert grams to kilograms

\[245_{grams}*(\frac{ kg }{ 1000grams }) = 0.245 kg \]

putting this together for molality

\[\frac{ moles of glucose }{ kg of water } \]

once you find molality all you need to do is multiply the molality by the constant that is given and you get the freezing point depression.

tell me what you get for the answer.

Answer 5

\[ik_{f}m = \Delta T \]

Answer 6

@sweetburger

Answer 7

@amistre64

Answer 8

@Abmon98

Answer 9

@Abmon98