45.2 g coconut oil containing
Answer:
The freezing point of the solution is -1.4°C
Explanation:
Freezing point decreases by the addition of a solute to the original solvent, <em>freezing point depression formula is:</em>
ΔT = kf×m×i
<em>Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.</em>
Molality of the solution is:
-moles CaBr₂ (Molar mass:
189.9g ₓ (1mol / 199.89g) = 0.95 moles
Molality is:
0.95 moles CaBr₂ / 3.75kg water = <em>0.253m</em>
Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:
CaBr₂(s) → Ca²⁺ + 2Br⁻
3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.
Replacing:
ΔT = kf×m×i
ΔT = 1.86°C/m×0.253m×3
ΔT = 1.4°C
The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,
<h3>The freezing point of the solution is -1.4°C</h3>
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Answer: Kc= 4.653
Explanation:
From The equation of reaction
CO(g)+2H2(g)⇌CH3OH(g)
From the question
[CH3OH] =0.0679 moles
[CO]= 0.160moles
[H2]= 0.302
Kc = [CH3OH]/[CO] [H2]^2
Kc=[0.069]/[0.16]×[0.302]^2
Kc= 4.653
