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5. When 4.50 g of a substance are dissolved in 125 g of CCl, (M-153.8 g/mol), the freezing point of the solution is depressed by 3.9 K. KCCl) 30 and K,(CC4) 4.95 K kg mol 0.64 K] a) What is the boiling point elevation of the solution? b) By what fraction is the vapor pressure of the solvent reduced? 1

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Answer #1

mass of the substance = 4.5g

depression in freezing point = i * kf * molality of the solution

3.9 = i * 30 * (4.5/153.8*1000/125)

i = vant hoff factor = 0.555

elevation in boiling point = i * kb * molality of the solution

(Tb - T0b) = i * Kb * m

Elevation in boiling point = 0.555*4.95*(4.5/153.8*1000/125)

Elevation in boiling point = 0.643 K

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