Question

Ethanol has a Kb of 1.22 Degrees C/m and usually boils at 78.4 Degrees Celcius. How many mol of an nonionizing solute would need to be added to 47.84 g ethanol in order to raise the boiling point to 86.30?

Answer 1

0.3097 moles of an nonionizing solute would need to be added.

Step-by-step explanation:

Molal elevation constant =
`k_b=1.22^oC/m`

Normal boiling point of ethanol =
`T_o=78.4^oC`

Boiling of solution =
`T_b=86.30^oC`

Moles of nonionizing solute = n

Mass of ethanol (solvent) = 47.84 g

Elevation boiling point:

`\Delta T_b=T_b-T_o`

`\Delta T_b=86.30^oC-78.4^oC=7.9^oC`

`\Delta T_b=K_b*  m`

`m=\frac{\text{Moloes of solute}}{\text{Mass of solvent(kg)}}`

`7.9^oC=1.22^oC/m* (n)/(0.04784 kg)`

n = 0.3097 mol

0.3097 moles of an nonionizing solute would need to be added.