compound with the empirical formula C8H5 is dissolved in benzene at 25°C. When 1

Question

compound with the empirical formula C8H5 is dissolved in benzene at 25°C. When 10.0 g of the compound are dissolved in 100 g of benzene, the vapor pressure above the resulting solution is 91.64 mm Hg. If the vapor pressure of pure benzene is 95.18 mm Hg at 25°C, calculate the molecular weight and determine the molecular formula of the compound. You may assume the compound is nonvolatile and that benzene obeys Raoult’s law in the solution. compound with the empirical formula C8H5 is dissolved in benzene at 25°C. When 10.0 g of the compound are dissolved in 100 g of benzene, the vapor pressure above the resulting solution is 91.64 mm Hg. If the vapor pressure of pure benzene is 95.18 mm Hg at 25°C, calculate the molecular weight and determine the molecular formula of the compound. You may assume the compound is nonvolatile and that benzene obeys Raoult’s law in the solution. compound with the empirical formula C8H5 is dissolved in benzene at 25°C. When 10.0 g of the compound are dissolved in 100 g of benzene, the vapor pressure above the resulting solution is 91.64 mm Hg. If the vapor pressure of pure benzene is 95.18 mm Hg at 25°C, calculate the molecular weight and determine the molecular formula of the compound. You may assume the compound is nonvolatile and that benzene obeys Raoult’s law in the solution.

Explanation / Answer

The lowering of vapour pressure is a colligative property

Vapour pressure of solution = vapour pressure of pure solvent X mole fraction of solvent

91.64 = 95.18 X mole fraction of benzene

0.962 = mole fraction of benzene

The mass of benzene = 100grams

Mol wt of benzene = 78 g

So moles of benzene = mass / mol wt = 100 / 78 = 1.28 moles

Mole fraction = moles of benzene / moles of benzene + moles of solute

0.962 = 1.28 / 1.28 + moles of solute

1.23 + moles of solute X 0.962 = 1.28

Moles of solute = 0.05 / 0.962 = 0.052 moles

molecular weight of given compound = mass of compound / mol wt of compound

0.052 = 10 / mol wt

mol wt of compound = 192.30

The empirical formula of the compound = C8H5

Let the molecular formula of the compound is C8xH5x

so

12 X 8 X x + 5x = 192.30

101x = 192.3

x = 1.9

So the probalble molecular formula will be

C16H10

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