Question

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.)

Answer 1

Answer: 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 %

Step-by-step explanation:

• Calculating the molarity of solution:

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

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

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:

`\text{Molarity of camphor}=(70* 1000)/(152.2* 575)\\\\\text{Molarity of camphor}=0.799M`

• Calculating the molarity of solution:

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

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

`0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL* 575mL)=451.38g`

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

`\text{Molality of solution}=\frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}`

where,

`m_(solute)` = Given mass of solute (camphor) = 70 g

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

`W_(solvent)` = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

`\text{Molality of camphor}=(70* 1000)/(152.2* 451.38)\\\\\text{Molality of camphor}=1.02m`

• Calculating the mole fraction of camphor:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

For camphor:

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

`\text{Moles of camphor}=(70g)/(152.2g/mol)=0.459mol`

For ethanol:

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

`\text{Moles of ethanol}=(451.38g)/(46g/mol)=9.813mol`

Mole fraction of a substance is given by:

`\chi_A=(n_A)/(n_A+n_B)`

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

`\chi_((camphor))=(0.459)/(10.272)=0.045`\

• Calculating the mass percent of camphor:

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

`\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}* 100`

Mass of camphor = 70 g

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

Putting values in above equation, we get:

`\text{Mass percent of camphor}=(70g)/(521.38g)* 100=13.43\%`

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 %