Question

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Answer 1

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ 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

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.