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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

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5 0

4 years ago

The volume occupied by 1.5 mole of gas at 35°C and 2.0 atmosphere of pressure is blank liters

Aneli [31]

Answer: 18.97 L

Explanation:

This can be solved by the Ideal Gas equation:

$P.V=n.R.T$

Where:

$P=2 atm$ is the pressure of the gas

$V$ is the volume of the gas

$n=1.5 mole$ the number of moles of gas

$R=0.0821\frac{L.atm}{mol.K}$ is the gas constant

$T=35\°C+273.15=308.15 K$ is the absolute temperature of the gas in Kelvin

Finding $V$ :

$V=\frac{nRT}{P}$

$V=\frac{(1.5 mole)(0.0821\frac{L.atm}{mol.K})(308.15 K)}{2 atm}$

$V=18.97 L$

Therefore:

The volume occupied by 1.5 mole of gas at 35°C and 2.0 atmosphere of pressure is 18.97 liters

7 0

4 years ago

The mass of a hypothetical planet is 1 100 that of the earth and it’s radius is 1 4 that of earth. If a person weighs 500 n on e

avanturin [10]

Answer: Your answer is 24

Explanation:

6 0

2 years ago

Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib

KengaRu [80]

Answer:

For methanol: Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol: Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol: Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

For methanol:

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol:

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol:

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

5 0

3 years ago

What is the limiting reactant in

MArishka [77]

Answer:

The limiting reactant is the K

Explanation:

This is the reaction:

2 K + F₂ → 2KF

2 moles of potassium react with 1 mol of fluorine

Let's determine the moles:

6.02x10²³ molecules are contained in 1 mol

10 molecules are contained in (10 . 1) / NA = 1.66x10⁻²³ moles

For 1 mol of flourine, we need 2 moles of K

If we have 1.66x10⁻²³ moles of Fl₂, we will need the double of K

For 2 moles of K, we need 1 mol of Fl₂

If we have 1.66x10⁻²³ moles of K, we will need the half of Fl₂

Then, the limiting reactant is the K

5 0

3 years ago