What is the freezing point of a solution that contains 36.0 g of glucose in 500.0 g of water (kf for water is 1.86c/m. the molar mass of glucose is 180.0 g per mole.)
a 0.755
b -0.744
c 1.49
d-1.49
what is the molality of a solution of water and kcl if the boiling point of the solution is 103.07 degrees celcius. (kb for water =0.512 c/m, kcl is an ionic compound)
a 0.300
b 0.600
c 3.00
d 6.00
! i have been having so much trouble with this. have no idea where to even start!

1) The depression of the freezing point is a colligative property.

For non ionic compounds it may be calculated  with the formula:

ΔTf = Kf * m

Where Kf is the cryoscopic constant of the water = 1.86°C/mol

And m is the molality, m = number of moles of solute / kg of solvent

You have 500.0 g of solvent = 0.5000 kg

And you can calculate the moles in 36.0 g of glucosa dividing  by its molar mass:

number of moles = mass in grams / molar mass = 36.0 g / 180.0 g/mol =  0.2 mol

=> m = 0.2 mol / 0.5000 kg = 0.4 m

=> ΔT = 1.86°C/m * 0.4m = 0.744 °C.

Given that the freezing point of pure water is 0.0°C, the new freezing point is 0°C - 0.744 °C = - 0.744°C

- 0.744 °C

2)  Elevation of boiling point is another colligative property.

It is calculate as:

ΔTb = i * Kb * m.

Whre i is van't Hoff constant, which accounts for the number of ions generated in ionic compounds.

Here, you do not have the i constant, but you can tell that each molecule of KCl dissociates into 2  ions (one K+ and one Cl-) which is to say that i = 2.

Kb is the boiling constant of the solvent, which here is water => Kb = 0.512 °C/m.

You also know the boiling point of the solution as 103.07°C. Given that the normal boiling point of pure water is 100.0°C, ΔT is 103.07°C - 100.0 °C = 3.07°C.

Then you can determine the molality, m = ΔTb / (i*Kb) = 3.07°C / (2*0.512°C/m) = 3.00 °C.

3.0 °C

D.na is the answer hope this