Question

What is the boiling point of a solution of 12.0 g of urea in 165.0 g of water

Answer 1

Answer:

T = 100.63 °C

Explanation:

To solve this question, we need to know what are we talking about here. In this case, we want to know the boiling point of a solution with Urea in water. This is a colligative property, so, the expression to use to calculate that is the following:

ΔT = m * K / MM * kg water (1)

Where:

ΔT: difference of temperatures (Tb of solution - Tb water)

m: mass of the urea

K: ebulloscopic constant of the water (0.52 ° C / m)

MM: molecular mass of urea

The boiling point of water is 100 °C, we have the mass of the urea, but not the molar mass. The urea has the formula CH₄N₂O, so the molar mass can be calculated using the atomic mass of the elements (I will use a rounded number for this):

MM = 12 + (4*1) + (2*14) + 16 = 60 g/mol

Now, we can calculate the ΔT and then, the boiling point of the solution:

ΔT = 12 * 0.52 / 60 * 0.165

ΔT = 6.24 / 9.9

ΔT = 0.63 °C

the value of ΔT is a difference between the boling point of water and the solution so:

ΔT = Ts - Tw

Ts = ΔT + Tw

Replacing we have:

Ts = 100 + 0.63

Ts = 100.63 ° C