What is the freezing point of an aqueous solution that boils at 105.0 ∘C? Express your answer using two significant figures.
Answers
The boiling point of standard water is 100 degree Celsius, with the addition of solute the boiling point is elevated. The freezing point of the solution will be -18.04 degree Celsius.
What is boiling point?
The boiling point is the temperature at which the liquid is converted to vapor. The change in boiling point of the aqueous solution gives the molality of the solution as:
The depression in freezing point from molality is given as;
The freezing point of aqueous water is zero degree Celsius. The freezing point of the solution will be:
The freezing point of the solution is -18.04 degree Celsius.
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Answer:
T°fussion of solution is -18°C
Explanation:
We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1
First of all, we apply boiling point elevation
ΔT = Kb . m . i
ΔT = Boiling T° of solution - Boiling T° of pure solvent
Kb = ebuliloscopic constant
105°C - 100° = 0.512 °C kg/mol . m . 1
5°C / 0.512 °C mol/kg = m
9.7 mol/kg = m
Now that we have the molality we can apply, the Freezing point depression.
ΔT = Kf . m . i
Kf = cryoscopic constant
0° - (T°fussion of solution) = 1.86 °C/m . 9.76 m . 1
- (1.86°C /m . 9.7 m) = T°fussion of solution
- 18°C = T°fussion of solution
Related Questions
In the chemical reaction in which sucrose is heated and decomposes to form carbon dioxide and water, which of the following is a reactant?a. sucrose c. water b. carbon dioxide d. heat
"Henry mixed salt and water together in a cup until he observed a clear solution. He measured the mass of the solution. Then he placed the cup outside for several sunny days during the summer. After a week, he observed that only solid salt remained in the cup and the mass had decreased. Henry concluded that a physical and chemical change occurred in this investigation."Which statements correctly defend or dispute his conclusion?A.)He is correct. Dissolving salt in water is a physical change, but evaporating the water is a chemical change. Formation of a solid is evidence that a chemical change occurred.B.)He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.C.)He is incorrect. Dissolving salt in water and evaporation of the water are both physical changes. The reappearance of salt is evidence that the change was reversible by a physical change, so it could not be a chemical change.D.)He is incorrect. Dissolving salt in water and evaporation of the water are both chemical changes. The reappearance of salt is evidence that the change was reversible by a chemical change, so it could not be a physical change.
Calculate the pH of a solution that is 0.111 M HCl. PH=?
Which group of elements ,in the periodic table, would most likely be used for electrical wires?
(2) Oxidation occurs as electrons are lost at the anode.(3) Reduction occurs as electrons are gained at the anode.
(4) Reduction occurs as electrons are lost at the cathode.
Answers
Oxidation occurs as electrons are lost at the anode at one of the electrodes in both an electrolytic cell and a voltaic cell. The answer is number 2. The rest of the choices does not answer the question given above.
Answers
Answer:
(3) saturated
Explanation:
Hello,
In this case, a solution that is at equilibrium must be (3) saturated as at the equilibrium, the entire solute is completely dissolved into the solvent, so no leftovers are present. Unsaturated solutions are not at equilibrium since there is a portion of solvent that is not contributing to equilibrium. Concentrated solutions are not at equilibrium since they could have more solution than allowed into the solvent.
Best regards.
Answers
Matter is made up of very small particles called Atom.
What is an Atom ?
An atom is the smallest unit of ordinary matter that forms a chemical element.
Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small, typically around 100 picometers across.
An atom consists of a central nucleus that is usually surrounded by one or more electrons. Each electron is negatively charged. The nucleus is positively charged, and contains one or more relatively heavy particles known as protons and neutrons.
An atom is a particle of matter that uniquely defines a chemical element.
Therefore, Matter is made up of very small particles called Atom.
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An atom are the basic building blocks of ordinary matter. Atoms can join to form molecules, which in turn from most of the objects around you.
Atoms are composed of particles called protons, electrons, and neutrons.
Protons carry a positive electrical charge.
Electrons carry a negative electrical charge.
Neutrons carry no electrical charge at all.
Hope i helped! And i know you didn't ask for all that, but i put by the way. If you need anything else ask me.
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I hopes this helps
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Answers
Answer:
-196 kJ
Explanation:
By the Hess' Law, the enthalpy of a global reaction is the sum of the enthalpies of the steps reactions. If the reaction is multiplied by a constant, the value of the enthalpy must be multiplied by the same constant, and if the reaction is inverted, the signal of the enthalpy must be inverted too.
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
S(s) + O₂(g) → SO₂(g) ΔH = -297 kJ (inverted and multiplied by 2)
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
2SO₂(g) → 2S(s) + 2O₂(g) ΔH = +594 kJ
-------------------------------------------------------------
2S(s) + 3O₂(g) + 2SO₂(g) → 2SO₃(g) + 2S(s) + 2O₂(g)
Simplifing the compounds that are in both sides (bolded):
2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = -790 + 594 = -196 kJ
Final answer:
The enthalpy of the reaction where sulfur dioxide is oxidized to sulfur trioxide is -395 kJ.
Explanation:
The calculation of the enthalpy change of the reaction in which sulfur dioxide is oxidized to sulfur trioxide involves Hess's Law, which states that the enthalpy change of a chemical reaction is the same whether it takes place in one step or several steps. This can be solved by comparing the enthalpy changes given in the two reactions presented.
First, consider the reactions given:
2S(s) + 3O₂(g) → 2SO₃(g), ΔH = -790 kJ
S(s) + O₂(g) → SO₂(g), ΔH = -297 kJ
From these reactions, it is seen that the first reaction can be re-written as:
2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = -790 kJ
However, this reaction contains two moles of SO₂ whereas the reaction in question only requires one mole. Thus, the enthalpy change for the reaction becomes: ΔH = -790 KJ / 2 = -395 kJ.
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