1. Will the boiling point of a solution of 10 g of NaCl (which has no appreciabl

Question

1. Will the boiling point of a solution of 10 g of NaCl (which has no appreciable vapor pressure at 100 oC) in 100 mL of water be higher than, lower than, or the same as the boiling point of water? Explain.

2. At 90 oC, the vapor pressures of benzene and toluene are, respectively, 1010 and 405 mm Hg.

Calculate the vapor pressure of a solution of 20 g of benzene (C6H6) and 20 g of toluene (C7H8) at 90 oC. Is the boiling point of this mixture higher or lower than 90 oC at an atmospheric pressure of 1 atm? Show all work.

Explanation / Answer

1) The boiling point of the solution will be higher than that of pure water as NaCl is non-volatile, it reduces the surface are for water molecules to evaporate and thus, the B.P of the solution gets higher.

2) Molar mass of benzene = 78 g/mole

Molar mass of toluene = 92 g/mole

Thus, moles of benzene in 20 g of it = mass/molar mass = 20/78 = 0.256

moles of toluene in 20 g of it = mass/molar mass = 20/92 = 0.217

mole fraction of benzene = moles of benzene/(moles of benzene+moles of toluene) = 0.541

Thus, mole fraction of toluene = 1- mole fraction of benzene = 0.459

Now, Vapor pressure of the mixture = P0benzene*mole fraction of benzene + P0toluene*mole fraction of toluene

or, Vapor pressure of the mixture = 1010*0.541 + 405*0.459 = 732.305 mm Hg

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