What is the boiling point of a solution of 12.0 g of urea (CH₄N₂O) in 165.0 g of water (K_b (H₂O) = 0.52 ^$\circ$C/m) A) 99.4^$\circ$C B) 100.0 ^$\circ$C C) 100.6^$\circ$C D) 106.2^$\circ$C E) 108.4^$\circ$C

Question

Quizplus · Accepted Answer

We can use the formula:

ΔT = Kf·m

where ΔT is the boiling point elevation, Kf is the freezing point depression constant (0.52°C/m for water), and m is the molality of the solution (moles of solute per kilogram of solvent).

First, we need to calculate the molality of the solution:

moles of urea = 12.0 g / 60.06 g/mol = 0.1998 mol
mass of water = 165.0 g = 0.165 kg
m = 0.1998 mol / 0.165 kg = 1.2109 m

Now we can calculate the boiling point elevation:

ΔT = Kf·m = 0.52°C/m · 1.2109 m = 0.631°C

Finally, we add this to the normal boiling point of water (100°C) to get the boiling point of the solution:

boiling point = 100°C + 0.631°C = 100.631°C

Rounding to one decimal place, we get:

Answer: C (100.6°C)

What Is the Boiling Point of a Solution of 12 ∘\circ∘

What Is the Boiling Point of a Solution of 12 $\circ$