I believe that it is petroleum ether.
Answer:
1. Caffeine, C₈H₁₀N₄O₂
Amount = 1.00/194 = 0.00515 moles
2. Ethanol, C₂H₅OH
Amount = 0.0217 moles
3. Dry Ice, CO₂
amount = 0.0227 moles
<em>Note: The question is incomplete. The compound are as follows:</em>
<em> 1. Caffeine, C₈H₁₀N₄O₂;</em>
<em>2. Ethanol, C₂H₅OH;</em>
<em>3. Dry Ice, CO₂</em>
Explanation:
Amount (moles) = mass in grams /molar mass in grams per mole
1. Caffeine, C₈H₁₀N₄O₂
molar mass of caffeine = 194 g/mol
Amount = 1.00 g/194 g/mol = 0.00515 moles
2. Ethanol, C₂H₅OH
molar mass of ethanol = 46 g/mol
Amount = 1.00 g/46 g/mol = 0.0217 moles
3. Dry Ice, CO₂
molar mass of dry ice = 44 g/mol
amount = 1.00 g/44 g/mol = 0.0227 moles
Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
D has a total of four significant figures.
