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

Answer:

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

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:

$\Delta T_b=T_b-T_o$

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

$\Delta T_b=K_b\times m$

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

$7.9^oC=1.22^oC/m\times \frac{n}{0.04784 kg}$

n = 0.3097 mol

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