What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of
1 answer:
73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.
Explanation:
Data given:
number of moles of CHCl3 = 1.31 moles
mass of solvent CHCl3 = 530 grams or 0.53 kg
Kf = 29.8 degrees C/m
freezing point of pure solvent or CCl4 = -22.9 degrees
freezing point = ?
The formula used to calculate the freezing point of the mixture is
ΔT = iKf.m
m= molality
molality =
putting the value in the equation:
molality=
= 2.47 M
Putting the values in freezing point equation
ΔT = 1.31 x 29.8 x 2.47
ΔT = 73.606 degrees
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Explanation:
The given reaction equation will be as follows.
Let is assume that at equilibrium the concentrations of given species are as follows.
M
M
M
Now, first calculate the value of as follows.
=
=
Now, according to the concentration values at the re-established equilibrium the value for will be calculated as follows.
M
Thus, we can conclude that the concentration of in the new equilibrium mixture is
M.
0.06105 moles is the number of moles of water lost.
<h3>What is molar mass?</h3>Molar mass is defined as the mass in grams of one mole of a substance.
Given data:
Mass of water = 1.1g
Molar mass water = 18.016 g/mol
Moles of water =?
These quantities are related by the following equation;
Moles =
Substituting the values of the quantities and solving for moles, we have;
Moles = = 0.06105 moles.
Hence, the 0.06105 moles is the number of moles of water lost.
Learn more about molar mass here:
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0.042 moles of Hydrogen evolved
<h3>Further explanation</h3>Given
I = 1.5 A
t = 1.5 hr = 5400 s
Required
Number of Hydrogen evolved
Solution
Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.
Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)
Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻
Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)
So at the cathode H₂ gas is produced
Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C
Q = i.t
Q = 1.5 x 5400
Q = 8100 C
mol e⁻ = 8100 : 96500 = 0.084
From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042