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.

Step-by-step 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 = (no. of moles)/(mass of solvent (in kg))\\= (0.793 mol)/(0.758 kg)\\= 1.05 m`

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

`\Delta T_(b) = i* K_(b) * 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* K_(b) * m\\3.80^(o)C = 1 * K_(b) * 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.