# What is the boiling point of a solution containing 2.33 g of caffeine, C_8H_10N_4O_2, dissolved in 15.0 g of benzene? The boiling point of pure benzene is 80.1 °C and the boiling point elevation constant, Kbp, is 2.53 °C/m.

Mar 26, 2016

The boiling point of the solution is 82.1 °C.

#### Explanation:

The formulae for boiling point elevation is

color(blue)(|bar(ul(ΔT_"b" = K_"b"m)|)

where

• ΔT_"b" is the boiling point elevation
• ${K}_{\text{b}}$ is the boiling point elevation constant
• $m$ is the molality of the solution

Our first task is to calculate the molality of the solution.

The molar mass of caffeine, $\text{Caf}$, is 194.19 g/mol.

$\text{moles of Caf" = 2.33 color(red)(cancel(color(black)("g Caf"))) × "1 mol Caf"/(194.19 color(red)(cancel(color(black)("g Caf")))) = "0.012 00 mol Caf}$

$\text{Molality" = "moles of solute"/"kilograms of solvent" = "0.012 00 mol"/"0.0150 kg" = "0.800 mol/kg}$

Now, we calculate the boiling point elevation.

ΔT_"b" = K_"b"m = "2.53 °C·"color(red)(cancel(color(black)("kg·mol"^"-1"))) × 0.800 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "2.02 °C"

Finally, we calculate the new boiling point.

${T}_{\text{b" = T_"b"^° + ΔT_"b" = "80.1 °C + 2.02 °C" = "82.1 °C}}$