Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you dissolve 70.0 g of camphor (C10H16O) in 575 mL of ethanol, C2H5OH. Calculate the molarity, molality, mole fraction, and weight percentage of camphor in this solution. (The density of ethanol is 0.785 g/mL.)

The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

Explanation:

To calculate the molarity of solution, we use the equation:

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

To calculate the mass of ethanol, we use the equation:

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

To calculate the molality of solution, we use the equation:

where,

= Given mass of solute (camphor) = 70 g

= Molar mass of solute (camphor) = 152.2  g/mol

= Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

To calculate the number of moles, we use the equation:

    .....(1)

For camphor:

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

For ethanol:

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

Mole fraction of a substance is given by:

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

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To calculate the mass percentage of camphor in solution, we use the equation:

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

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