What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2

(formula weight = 164 g/mol) in 115 g of water? The molal freezing point depression constant for
water is 1.86 °C/m.

whats wrong with my answer:

molal: 0.069 moles Ca(NO3)2/0.115kg= 0.6
0.6x1.86=1.116
but its wrong, the answer is supposed to be -3.34

You forgot that Ca(NO3)2 ionizes into three ions so the van't Hoff factor is 3.

delta T = i*Kf*m
So multiply your answer by 3 and you'll have it.

why is it negative?

bc adding solute to soln decreases the freezing pt.

Oh, looks like your calculations took a wrong turn! Let's break it down in a more entertaining way.

Hocus-pocus, let's solve this freezing point mystery! You dissolved 11.3 grams of Ca(NO3)2 in 115 grams of water. The formula weight of Ca(NO3)2 is 164 g/mol, informing us that you've got approximately 0.069 moles of Ca(NO3)2.

Now, to determine the molal freezing point depression, we need to divide the molality (moles of solute per kilogram of solvent). So, it's 0.069 moles divided by 0.115 kilograms, giving us a molality of about 0.6 "moles per kilogram."

But wait, there's more! We need to multiply this molality by the freezing point depression constant of water, which is 1.86 °C/m. So, 0.6 times 1.86 equals 1.116 °C.

Seems like you were tricked by an optical illusion! The right answer should be -1.116 °C, not -3.34 °C. Keep in mind that the sign indicates a temperature decrease, not a drop in mathematical skills.

So, wave your wand again and give it another shot!

Your calculation for the molality (molal) is correct, which is 0.6 mol/kg. However, the next step in your calculation is incorrect. To find the freezing point depression, you need to multiply the molality by the molal freezing point depression constant for water (1.86 °C/m). So the correct calculation is:

Freezing point depression = molality x molal freezing point depression constant

Freezing point depression = 0.6 mol/kg x 1.86 °C/m
Freezing point depression = 1.116 °C

Now, to find the freezing point of the solution, you need to subtract the freezing point depression from the freezing point of pure water. The freezing point of pure water is 0 °C. Therefore, the freezing point of the solution is:

Freezing point = Freezing point of water - Freezing point depression
Freezing point = 0 °C - 1.116 °C
Freezing point = -1.116 °C

However, since the question asks for the freezing point in °C, it should be rounded to two decimal places:

Freezing point ≈ -1.12 °C

So the correct freezing point of the solution prepared by dissolving 11.3 g of Ca(NO3)2 in 115 g of water is approximately -1.12 °C.

It seems that there might have been a calculation error in your working, resulting in a different value. Double-checking your calculations and using the correct formula and constant should lead you to the correct answer.