Respuesta :
The change in the freezing point of water when 35.5 g of sucrose is dissolved in 55.0 g of water is - 3.51°C .
What is freezing point depression ?
When a solute is added to a pure solvent, the freezing point drops. This is known as freezing point depression
It is a colligative property, which means that the effect is proportional to the number of solute particles introduced, regardless of the molecule.
The equation used to calculate the freezing point depression is
ΔT = i[tex]\rm K_f[/tex] m
where ,
ΔT is the change in freezing point,
i is the van't Hoff factor,
[tex]\rm K_f[/tex] is the freezing point depression constant, and
m is the molality of the solution.
It is given in the question that the
[tex]\rm K_f[/tex] of water = -1.86°C/mol
molar mass sucrose = 342.30 g/mol
i value of sugar = 1
35.5 g of sucrose is dissolved in 55.0 g of water
ΔT = 1 * -1.86° *m
molality = (35.5/342.3)/(55/1000)
m = 1.88
ΔT = 1 * -1.86° * 1.88
ΔT = - 3.51°C
The change in the freezing point of water when 35.5 g of sucrose is dissolved in 55.0 g of water is - 3.51°C .
To know more about freezing point depression
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