Answer: The value of [tex]K_{b}[/tex] for chloroform is [tex]3.62^{o}C/m[/tex] when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
Explanation:
Given: Moles of solute = 0.793 mol
Mass of solvent = 0.758
[tex]\Delta T_{b} = 3.80^{o}C[/tex]
As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.
[tex]Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m[/tex]
Now, the values of [tex]K_b[/tex] is calculated as follows.
[tex]\Delta T_{b} = i\times K_{b} \times m[/tex]
where,
i = Van't Hoff factor = 1 (for chloroform)
m = molality
[tex]K_{b}[/tex] = molal boiling point elevation constant
Substitute the values into above formula as follows.
[tex]\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m[/tex]
Thus, we can conclude that the value of [tex]K_{b}[/tex] for chloroform is [tex]3.62^{o}C/m[/tex] when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
