Question

# When 8.05 g of an unknown compound X was dissolved in 100 g of benzene C6H6,...

When 8.05 g of an unknown compound X was dissolved in 100 g of benzene C6H6, the vapor pressure of the benzene decreased from 100 torr to 94.8 torr at 26 degrees C. (MW C6H6 =78.0 g/mol.)

a.) What is the mole fraction for unknown compound X?

b.) What is the molar mass of unkown compound X?

#### Homework Answers

Answer #1

relative lowering vapour is equal to mole fraction of solute

Po -Ps / Po = mole fraction of unknown

100 - 94.8 / 100 = mole fraction of unknown

0.052 = mole fraction of unknown

a ) mole fraction of unknown = 0.052

b)

moles of benzene = n2 = 100 / 78 = 1.282

moles of unknown = n1

mole fraction of unknown = n1 / n1+ n2

0.052 = n1 / 1.282 + n1

0.067 + 0.052 n1 = n1

n1 = 0.0707

moles of unknown = 0.0707

8.09 / molar mass = 0.0707

molar mass = 114.4 g/mol

molar mass of unkown compound X = 114.4 g/mol

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