What is the boiling point of a solution containing 0.200 mol of a nonvolatile solute in 125 g of benzene?

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Number of moles of solute = 0.200 mol

Weight of solvent (Benzene) = 125 g = 0.125 kg

Molality of solution "= \\frac{0.20}{0.125} = 1.6 \\;m"

Elevation of boiling point "= \u0394T_{bp} = K_{bp} m_{solute}"

"K_{bp}" = Molal elevation boiling point constant

"K_{bp} = 2.53 \\; \u00b0C\/m" for benzene

"\u0394T_{bp} = 2.53 \\times 1.6 = 4.048 \\;\u00b0C"

Boiling point of solution = Boiling point of pure solvent + Elevation in boiling point

= 80.1 °C + 4.048 °C = 84.15 °C

Boiling point of solution = 84.15 °C