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

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

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,

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

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:

$4.2^oC=1\times 1.22^oC/m\times \frac{m_{solute}\times 1000}{132.15g/mol\times 175}\\\\m_{solute}=79.6g$

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