Answer:
Freezing point of the solution is -1.07°C
Explanation:
The addition of a solute to a solvent decreases its freezing point following the formula:
ΔT = Kf×m×i
Where ΔT is change in temperature (0°C - X); Kf is constant of solvent (1.86°C/m), m is molality (moles glucose per kg solvent) and i is Van't Hoff factor (1 for glucose).
Molality of the solution is:
15.5g glucose -Molar mass 180.156g/mol- contains:
15.5g × (1mol / 180.156g) = 0.086 moles of glucose / 0.150 kg water = 0.574 m
Replacing:
0°C - X = 1.86°C/m × 0.574 × 1
-X = 1.07°C
Freezing point of the solution is -1.07°C
The freezing point of an aqueous solution of 15.5 g glucose dissolved in 150 g water will be "-1.07°C".
According to the question,
Constant of solvent, [tex]K_f[/tex] = 1.86°C/m
Change in temperature, ΔT = 0°C - X
Grams of glucose = 15.5 g
Grams of water = 150 g
We know the relation,
Solution's molality = 15.5 × [tex]\frac{1 \ mol}{180.156}[/tex]
= [tex]\frac{0.086}{0.150}[/tex]
= 0.574 m
hence,
→ ΔT = [tex]K_f[/tex] × m × i
By substituting the values,
0°C - X = 1.86° × 0.574 × 1
X = -1.07°C
Thus the answer above is correct.
Find out more information about molality here:
https://brainly.com/question/14623340
