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Suppose that 13g of a nonelectrolyte is dissolved in 0.50 kg of benzene. The
boiling point of this solution is 80.61 °C. The normal boiling point of benzene is
80.10 °C. The molal boiling point elevation constant for benzene is 2.53 °C/m. What is the
molar mass of the solute?

Respuesta :

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Answer:

128.98 g/Mol

Explanation:

ΔTb = Kb * m * i

Where;

ΔTb = boiling point elevation

Kb = boiling point constant

m = molality of the solution

i = Van't Hoff factor

Hence;

Where molality = number of moles of solute/mass of solvent in Kg

Number of moles of solute = mass/Molar mass

So;

80.61 - 80.10 = 2.53 * 13g/M/0.5 * 1

Where M is the molar mass of the solute

0.51 = 2.53 * 13g/M/0.5

0.51 = 2.53 * 13/M * 1/0.5

0.51 = 32.89/M * 1/0.5

0.51 * 0.5 = 32.89/M

0.255 M = 32.89

M = 32.89/0.255

M= 128.98 g/Mol

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