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:
0.367M
Explanation:
Molarity refers to the molar concentration of a solution. It can be calculated using the formula below:
Molarity = n/V
Where;
n = number of moles (mol)
V = volume (L)
According to the given information in this question;
n = 0.55 mole
V = 1.50 L
Molarity = 0.55/1.50
Molarity = 0.367M
Answer: Thus 0.724 mol of 
are needed to obtain 18.6 g of 
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of are produced by = 1 mole of
Thus 1.09 moles of will be produced by =
of
But as yield of reaction is 75.6 %, the amount of needed is =
Thus 0.724 mol of are needed to obtain 18.6 g of
Answer:
![K=\frac{[CaO][CH_{4}][H_{2}O ]^{2} }{[CaCO_{2}][H_{2}]^4 }](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCaO%5D%5BCH_%7B4%7D%5D%5BH_%7B2%7DO%20%5D%5E%7B2%7D%20%20%7D%7B%5BCaCO_%7B2%7D%5D%5BH_%7B2%7D%5D%5E4%20%20%7D)
Explanation:
The equilibrium expression is the K value equal to the product of the concentrations of the products over the product of the concentrations of the reactants. If there is a coefficient in front of the compound, raise the molecule to that power.
Since K is big, more product is expected. This is because of mathematic principles. A large numerator with a small denominator will produce a large number.
Answer:
Explanation:
Option D electrolysis is the correct answer
