Question

When 14.7g phosphorus is dissolved in .2 kg benzene the boiling point increases from 80.1*C to 81.6*C. Calculate the molar mass of the phosphorus. The benzene is solvent.

Answer 1

Kb of benzene = 2.53 oC / m (Literature value)

let molar mass of phosphorus = M g / mole

mole of phosphorus = (14.7 / M) mole

molality = (14.7 / M ) mole / 0.2 Kg = (73.5 / M) mole / Kg

we know,

delta Tb = i * Kb * m

or

(81.6 - 80.1) oC = 1 * 2.53 * (73.5 / M)

or

M = 123.97 g / mole

thus

molar mass of phosphorus = 123.97 g / mole