A solution contains 20.0 g of C6H12O6 (glucose) in 250 g. of water.A. what is freezing point depression of solvent?
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Answer:
The freezing point depression of solvent is 0.8265 K.
Explanation:
Mass of glucose = 250 g
Mass of solvent = 20.0 g = 0.020 kg
Molal depression water constant of water, = 1.86 K kg/mol
Molality of the solution =
Normal freezing point of water =
Freezing point of solution =
The freezing point depression of solvent is 0.8265 K.
number of moles:
20 / 180 = 0.111 moles
Δ t f * ( Kf) *( i ) * ( m )
250 g of water in Kg = 250 / 1000 = 0.25 kg
1.86 ºC/m * 0.111 / 0.25
1.86ºC/m * 0.444
= 0.82584ºC
Fp = 0ºC - 0.82584
Fp = - 0.82584ºC
hope this helps!
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Learn more:
1. The major contribution of Antoine Lavoisier to chemistry: brainly.com/question/2500879
2. Example of physical change: brainly.com/question/1119909
Answer details:
Grade: High School
Subject: Chemistry
Chapter: Atomic Structure
Keywords: Rutherford, atom, volume, nucleus, orbit, postulates, limitations, Maxwell, electromagnetic radiations, distribution of electrons, stability of atom.
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