All categories

4 years ago

What effect does increasing the concentration of a dissolved solute have on each of the colligative properties?

Chemistry

1 answer:

Igoryamba4 years ago

7 0

Answer:

If we increase the concentration of a dissolved solute, the solution would have a vapor pressure so much low, the boiling temperature for the solution will be so high, freezing point for the solution will be so much low and the osmotic pressure will be higher.

Colligative properties always depends on dissolved particles (solute)

Explanation:

These are the colligative properties

- Vapor pressure lowering

ΔP = P° . Xm

Vapor pressure of pure solvent - Vapor pressure of solution.

If we add more solute, it would raise the Xm, so the solution would have a vapor pressure so much low.

Vapor pressure pure solvent - Vapor pressure solution ↑ = P° . Xm ↑

- Boiling point elevation

ΔT = Kb . m

When we add more solute, we are increasing the molality.

↑T° boiling of solution - T° boiling pure solvent = Kf . m ↑

Boiling temperature for the solution will be so high.

- Freezing point depression

When we add more solute, we are increasing the molality.

ΔT = Kf . m

T° fussion of pure solvent - ↓T° fussion of solution = Kf . m↑

Freezing point for the solution will be so much low.

- Osmotic pressure

π = M . R . T

When we add solute, molarity is increasing. Therefore the osmotic pressure will be higher.

π↑ = M↑ . R . T

You might be interested in

A bowling ball, a baseball, a tennis ball, and a ping-pong ball are all dropped from a height of 2 meters. Which has the most ki

mr_godi [17]

Answer:

they all have the same amount of kinetic energy

6 0

3 years ago

A chemistry student is given the task of analyzing the three unknown samples in the previous question. Compare the densities fro

OLEGan [10]

Answer:

It’s B

Explanation:

6 0

3 years ago

Can someone help me? It needs to have a diagram that has arrows.

daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0

3 years ago

A substance made of metal atoms only?

SVEN [57.7K]

It has to be a metal, it can't be anything else.

5 0

3 years ago

Represent the decomposition of aluminum oxide using the same number of atoms of molecules as are

pantera1 [17]

Answer:

Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.

Explanation:

<u>Decomposition reaction:</u>

When a single compound break down into two or more simpler products.

For example "AB" reactant undergoes decomposition to form "A" and "B" products.

The chemical reaction is as follows.

$AB\rightarrow A+B$

The given compound is aluminium oxide.

The decomposition reaction of aluminium oxide is a follows.

$Al_{2}O_{3}\rightarrow Al+O_{2}$

The balanced equation is as follows.

$2Al_{2}O_{3}\rightarrow 4Al+3O_{2}$

Therefore, Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.

4 0

3 years ago