Respuesta :
The question is: What is the boiling point of an aqueous solution of urea at 20% m / m (20g of solute for every 80g of solvent), if the molar mass of urea is 60 g / mol. (Kb = 0.52 ° C / m)
Answer: The boiling point of given aqueous solution of urea is [tex]2.145^{o}C[/tex].
Explanation:
Given: Mass of solute = 20 g
Mass of solvent = 80 g (1 g = 0.001 kg) = 0.08 kg
[tex]K_{b} = 0.52^{o}C/m[/tex]
Molar mass of urea = 60 g
Molality is the number of moles of solute present in a kg of solvent.
Moles is the mass of a substance divided by its molar mass. So, moles of urea is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{20 g}{60 g/mol}\\= 0.33 mol[/tex]
Now, molality of given solution is as follows.
[tex]Molality = \frac{moles}{mass(in kg)}\\= \frac{0.33 mol}{0.08 kg}\\= 4.125 m[/tex]
Formula used to calculate the boiling point is as follows.
[tex]\Delta T_{b} = K_{b}m\\= 0.52^{o}C/m \times 4.125 m\\= 2.145^{o}C[/tex]
Thus, we can conclude the the boiling point of given aqueous solution of urea is [tex]2.145^{o}C[/tex].
