Answer :
- Boiling point of the sugar solution will be higher than that of water's boling point.
- Freezing point of the sugar solution will be lower than that of water's freezing point.
Explanation:
- Boiling point of a liquid is defined as temperature at which vapor pressure of liquid becomes equal to the atmospheric pressure.
Boiling point of solution is always higher than that of the pure solvent
Vapor pressure increases with increase in temperature which means sugar solution will be heated more to make vapor pressure equal to atmospheric pressure.
- Freezing point is defined as temperature at which solid and liquid phase are at equilibrium or temperature at which vapor pressure of liquid becomes equal to the vapor pressure in its solid phase.
Freezing point of solution is always lower than that of the pure solvent.
Lower the temperature, lower will be the vapor pressure which sugar solution solution will get freeze at lower temperature than that of the water.
Answer:
I think option (d) is right answer
There would be a 2 infront of the NH3 and and 2 Infront of the H2O
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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