Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907 m sugar solution.

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.

m = molality of solute (m or )

Given : kf = 1.86

m = 0.907 )

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 * 0.907 )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

specific gravity is the density of a substance divided by the density of water. since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density

so "c" is the answer

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