Question

Cinnamaldehyde (MM = 132.15 g/mol) is used as a flavoring agent. What mass of cinnamaldehyde must be added to 175 g of ethanol to give a solution whose boiling point is 82.7°C? Kb = 1.22°C/m, boiling point of pure ethanol = 78.5°C

Answer 1

The mass of cinnamaldehyde required to increase the boiling point of 175 g of ethanol to 82.7°C is 79.453 grams. This is calculated using the molality and the boiling point elevation formula with the given constants.

Step-by-step explanation:

Using the boiling point elevation equation ΔT = Kb*m, where ΔT is the change in boiling point, Kb is the ebullioscopic constant, and m is the molality, we can find the molality required to achieve the desired boiling point elevation. The change in boiling point (ΔT) is the difference between the final boiling point (82.7°C) and the boiling point of pure ethanol (78.5°C), which is 4.2°C.

With Kb given as 1.22°C/m, we calculate the molality (m) of the solution:

m = ΔT / Kb = 4.2°C / 1.22°C/m = 3.44 m.

Next, we convert the mass of ethanol to kilograms to use in the molality formula:

175 g ethanol = 0.175 kg ethanol

Now, we can find the moles of cinnamaldehyde required:

moles of cinnamaldehyde = molality (m) × kilogram solvent = 3.44 m × 0.175 kg = 0.602 moles.

Finally, we can calculate the mass of cinnamaldehyde needed:

mass of cinnamaldehyde = moles × molar mass = 0.602 moles × 132.15 g/mol = 79.453 grams.

Therefore, to increase the boiling point of 175 g of ethanol to 82.7°C, 79.453 grams of cinnamaldehyde must be added.

Answer 2

Answer: The mass of cinnamaldehyde that must be added is 79.6 grams.

Step-by-step explanation:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

`\Delta T_b=\text{boiling point of solution}-\text{boiling point of pure solution}`

`\Delta T_b` = ? °C

Boiling point of pure ethanol = 78.5°C

Boiling point of solution = 82.7°C

Putting values in above equation, we get:

`\Delta T_b=(82.7-78.5)^oC=4.2^oC`

To calculate the elevation in boiling point, we use the equation:

`\Delta T_b=iK_bm`

Or,

`\Delta T_b=i* K_b* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}`

where,

`\Delta T_b` = 4.2°C

i = Vant hoff factor = 1 (For non-electrolytes)

`K_b` = molal boiling point elevation constant = 1.22°C/m.g

`m_(solute)` = Given mass of solute (cinnamaldehyde) = ? g

`M_(solute)` = Molar mass of solute (cinnamaldehyde) = 132.15 g/mol

`W_(solvent)` = Mass of solvent (ethanol) = 175 g

Putting values in above equation, we get:

`4.2^oC=1* 1.22^oC/m* (m_(solute)* 1000)/(132.15g/mol* 175)\\\\m_(solute)=79.6g`

Hence, the mass of cinnamaldehyde that must be added is 79.6 grams.