A solution contains naphthalene (c10h8) dissolved in hexane (c6h14) at a concentration of 10.00 % naphthalene by mass. part a ca
2 answers:
The vapor pressure at 25°C of hexane : <u>140.52 torr</u>
<h3>Further explanation </h3>Solution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.
Solution properties of electrolyte solutions differ from non-electrolyte solutions because electrolyte solutions contain a greater number of particles because electrolytes break down into ions. So the Solution properties of electrolytes is greater than non-electrolytes.
The term is used in the Solution properties
- 1. mole fraction
the ratio of the number of moles of solute to the mole of solution
- 2. Vapor pressure
Vapor pressure depends on the mole fraction of the components in the solution
P = Xs. P °
P = vapor pressure solution
P ° = pure vapor pressure of the solvent
Xs = mole fraction solvent
ΔP = P ° - P where
ΔP = change in vapor pressure
We determine the mole fraction of hexane (we consider a 100 g solution)
mass of naphthalene (C₁₀H₈) = 105 = 10 g
molar mass: 128 g/mole
mole: 10: 128 = 0.078
mass of hexane (C₆H₁₄) = 90 g
molar mass: 86 g/mole
mol = 90: 86 = 1.0465
Total mole = 1.0465 + 0.078 = 1.1245
Hexane mole fraction = 1.0465: 1.1245 = 0.9306
vapor pressure hexane = hexane mole fraction. vapor pressure of pure hexane
Phexane = 0.9306 x 151
Phexane = 140.52 torr
<h3>Learn more </h3>
colligative properties
Raoult's law
The vapor pressure of benzene
The freezing point of a solution
Answer : The vapor pressure of hexane = 140.5055 torr
Solution : Given,
Concentration of Naphthalene by mass = 10 %
Vapor pressure of pure Hexane = 151 torr
From the periodic table, the molar masses of Naphthalene and Hexane are:
Molar mass of Naphthalene = 128 g/mole
Molar mass of Hexane = 86 g/mole
From the Raoult's law, the vapor pressure of the component is equal to the vapor pressure of the pure component multiplied by its mole fraction.
Formula used :
..............(1)
Where,
= vapor pressure of component i
= vapor pressure of pure component i
= mole fraction of of component i in the solution
For this problem, few steps are involved:
step 1 : concentration of 10% naphthalene by mass means 10 g of naphthalene present in 100 g of solution.
This means,
The mass of naphthalene = 10 g
The mass of hexane = 100 - 10 = 90 g
step 2 : find the number of moles of naphthalene and hexane.
Formula used :
Number of moles =
Moles of Naphthalene = = 0.0781 moles
Moles of Hexane = = 1.0465 moles
Total moles = Moles of Naphthalene + Moles of Hexane
= 0.0781 + 1.0465 = 1.1246 moles
step 3 : Now, calculate mole fraction of Hexane
Mole fraction of Hexane = =
= 0.9305
step 4 : Now, find the vapor pressure of Hexane by using above formula (1).
Now put all the values in this formula, we get
Vapor pressure of Hexane = Vapor pressure of pure Hexane × Mole fraction of Hexane
= 151 torr × 0.9305 = 140.5055 torr
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The forward reaction rate would decrease if is removed from the mixture. The reason is that collisions between
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molecules would become less frequent.
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<h3>Equilibrium</h3>Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.
Whenever the forward reaction adds one mole of to the system, the backward reaction would have broken down the same amount of
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<h3>Collision Theory</h3>In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.
Assume that and
molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield
, only a fraction of the reactions will be fruitful.
Assume that molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.
Because fewer molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between
and
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The backward reaction rate is likely going to stay the same right after was taken out of the mixture without changing the temperature or pressure.
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As more and more gets converted to
and
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