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.
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I believe your answer will be 40
Answer:
Mass of seawater = 512.5 gram
Explanation:
Given:
Volume of seawater = 500 ml
Density of seawater = 1.025 g/ml
Find:
Mass of seawater
Computation:
Mass of seawater = Volume of seawater × Density of seawater
Mass of seawater = 500 × 1.025
Mass of seawater = 512.5 gram
Explanation:
Species present are
,
and
. Now, we will calculate the number of componenets present as follows.
C = n - E
where, E = number of independent equations
n = number of phases present
Therefore, we will calculate the number of components as follows.
C = 3 - 1
= 2
Since, sodium chloride (NaCl) will tend to form a homogeneous mixture in water. Therefore, phase will be equal to 1.
Therefore, it means that there are two components present in one phase.
Answer:
The temperature would be 194, 8 K.
Explanation:
We use the formula:
PV=nRT --> T=PV/nR
T= 0,500 atm x 28, 6 L/ 0,895 mol x 0,082 l atm /K mol
<em>T= 194,8494345 K</em>