Question

Since pure water boils at 100.00 ∘C , and since the addition of solute increases boiling point, the boiling point of an aqueous solution, Tb , will be

Tb=(100.00+ΔTb)∘C

Since pure water freezes at 0.00 ∘C , and since the addition of solute decreases freezing point, the freezing point of an aqueous solution, Tf , will be

Tf=(0.00−ΔTf)∘C

Part A

What is the boiling point of a solution made using 785 g of sucrose, C12H22O11 , in 0.250 kg of water, H2O ?

Express your answer to five significant figures and include the appropriate units.

Part B

What is the freezing point of a solution that contains 26.3 g of urea, CO(NH2)2 , in 245 mL water, H2O ? Assume a density of water of 1.00 g/mL .

Express your answer to three significant figures and include the appropriate units.

Answer 1

Part A

Given that mass of sucrose =785 g and water 0.250 kg

First calculate the moles of sucrose as follows:

moles sucrose = amount in g/ molar mass g/ moles

= (785 g / 342.3 g/mole) = 2.29 moles

molality = (2.29 moles / 0.250 kg)

= 9.17 m

ΔTb = kb * m = (0.512 * 9.17) = 3.7

bp = (bp of water + ΔTb)

bp = 109.17 C

part B

given that 26.3 g of urea, CO(NH2)2 , in 245 mL water, H2O

thus mass of water is 245 g

density = mass/ volume

mass = volume * density

= 245*1.0

= 245 g

Or 0.245 kg

moles urea = amount in g/ molar mass g/ moles

= (26.3 g / 60.06 g/mole) = 0.44 moles

molality = (0.44 moles / 0.245 kg)

= 1.79 m

ΔTb = kb * m = (1.86 * 1.79) = 3.3294 C

Tf =(0.00−ΔTf)∘C

= 0.00 -3.3294 )C

= 3.3294 C
bp = (bp of water + ΔTb)

bp = 109.17 C