Calculate the freezing point and boiling point of a solution containing 15.0 g o

Question

Calculate the freezing point and boiling point of a solution containing 15.0 g of naphthalene (C10H8) in 115.0 mL of benzene. Benzene has a density of .877 g/cm3. Calculate the freezing point of a solution (Kf(benzene) =5.12 degrees C/m) Calculate the boiling point of a solution (Kb(benzene) =2.53 degrees C/m)
Calculate the freezing point and boiling point of a solution containing 15.0 g of naphthalene (C10H8) in 115.0 mL of benzene. Benzene has a density of .877 g/cm3. Calculate the freezing point of a solution (Kf(benzene) =5.12 degrees C/m) Calculate the boiling point of a solution (Kb(benzene) =2.53 degrees C/m)
Calculate the freezing point of a solution (Kf(benzene) =5.12 degrees C/m) Calculate the boiling point of a solution (Kb(benzene) =2.53 degrees C/m)

Explanation / Answer

For calculating the freezing point:

Using relation:

dT = -Kf*m

Putting values:

dT = 5.12*((15/128)/(0.877*115/1000)) = 5.95

Freezing point of pure benzene = 5.50C

So,

Freezing point of solution = 5.5-5.95 = -0.450C

For calculating the boiling point:

Using relation:

dT = -Kb*m

Putting values:

dT = 2.53*((15/128)/(0.877*115/1000)) = 2.94

Freezing point of pure benzene = 80.10C

So,

Freezing point of solution = 80.1+2.94 = 83.040C

Hope this helps !

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