Respuesta :
Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.[tex] \frac{^0 C}{m} [/tex]
m = molality of solute (m or [tex] \frac{mol}{Kg} [/tex] )
Given : kf = 1.86 [tex] \frac{^0 C*Kg}{mol} [/tex]
m = 0.907 [tex] \frac{mol}{Kg} [/tex] )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 [tex] \frac{^0 C*Kg}{mol} [/tex] * 0.907[tex] \frac{mol}{Kg} [/tex] )
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C
