Respuesta :

BlueSky06

The solution would be like this for this specific problem:

For freezing point

Tf, soln= -114.6-(i*Kf*m) 
Tf, soln= -114.6 - (1*1.99*0.40) 
Tf= -115.4 C 

 

For boiling point:

Tb=78.4+(i*Kf*m) 

Tb=78.4+(1*1.99*.40) 
Tb= 79.2 C 

So the boiling and freezing point of a 0.40m solution of sucrose in ethanol is 79.2 C and  -115.4 C respectively.

ajinems

The freezing point of sucrose in ethanol is = 114.896°C

Calculation of freezing point of sucrose

The formula for freezing point is solved as follows:

Where,

ΔTf = T2(solution) - T1 (pure solvent)

Kf = cryoscopic constant(1.99 °C/m)

m = molality of solution(0.40M)

T1 = -114.1 °C (freezing point of ethanol)

Therefore, ΔTf= 1.99 × 0.40= - 0.796°C

But ΔTf= T2 - T1

- 0.796°C = T2 - (-114.1 °C)

- 0.796°C = T2 + 114.1 °C

To solve for T2 = -114.1 - 0.796

= −114.896

Therefore, the freezing point of sucrose = −114.896°C

Learn more about freezing point here:

https://brainly.com/question/24314907

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