Question

The boiling point of diethyl ether CH₃CH₂OCH₂CH₃ is 34.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is chlorophyll.

If 13.19 grams of chlorophyll, C₅₅H₇₂MgN₄O₅ (893.5 g/mol), are dissolved in 269.9 grams of diethyl ether.

The molality of the solution is  m.

The boiling point of the solution is  °C.

Answer 1

moles of chlorophyll = 13.19 / 893.5 = 0.014762  M

molality of the solution =   0.014762 / 0.269 = 0.054877 m

ΔT = iK_bm

where
ΔT = Change in temperature in °C
i = van 't Hoff factor
K_b = molal boiling point elevation constant in °C kg/mol
m = molality of the solute in mol solute/kg solvent.

molal boiling point elevation constant for diethyl ether = 2.16 °C kg/mol

SUbstitute in the equation and find out ΔT

ΔT = 1 x 2.16 x  0.054877= 0.11853

boiling point of the solution = 34.5 + 0.11853 = 34.62 °C