0.01 m
< 0.03 m
< 0.04 m urea
As molal concentration rises, so does freezing point depression. It can be expressed mathematically as ΔTf = Kfm.
<h3>What is Colligative Properties ?</h3>
- The concentration of solute particles in a solution, not the composition of the solute, determines a colligative properties .
- Osmotic pressure, boiling point elevation, freezing point depression, and vapor pressure reduction are examples of ligand-like properties.
<h3>What is freezing point depression?</h3>
- When less of another non-volatile material is added, the temperature at which a substance freezes decreases, a process known as Freezing-point depression.
- Examples include combining two solids together, such as contaminants in a finely powdered medicine, salt in water, alcohol in water.
- An significant factor in workplace safety is freezing points.
- If a substance is kept below its freezing point, it may become more or less dangerous.
- The freezing point additionally offers a crucial safety standard for evaluating the impacts of worker exposure to cold conditions.
Learn moree about Colligative Properties here:
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Answer:
a) 4869 kj will be released
b) 43.86 g of octane
Explanation:
The heat of combustion is the amount of heat released when one mole of a substance reacts with enough oxygen
since the heat of combustion is per mol of combustible substancewjat we are required to do in this problem is calculate number moles in the reactions although in a different manner
a) MW C3H6O = 158 g/ mol
mol C3H6O = 158 g × 1 mol/ 58.08 g
= 2.72 - 1790 kj / mol ×2.72 mol = 4869 kj
b) Here we are asked the mass of octane to produce 1950 kj of heat knowing that per mole of octane we get 5074.1 kj then
1 mol / 5074.1kj × 1950 kj= 0.384 mol
mass C8H18 = 0.384 mol × 114.23 g/ mol = 43.86 g
Answer: 3 oxygen atoms!
Explanation: Let me know if I'm wrong, if not I'm happy to help!
Multiply this number by avogadro's number- 6.02 times 10^23