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Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?
Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kg
Freezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.
Correct answer is '4'. Can you explain this answer?
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Molecular formula of white phosphorus in benzene is (P)x. What is the ...
w1/ m1 = moles of solute, w2 = mass of solvent
If molecular formula = (P)x
∴ Molar mass = 31x = 123.42
x = 4
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Molecular formula of white phosphorus in benzene is (P)x. What is the ...
Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on the following experiment?
Freezing point of benzene = 5.45°C
Kf (benzene) = 5.12 K mol⁻¹ kg
Freezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89°C
To determine the molecular formula of white phosphorus in benzene, we can use the concept of freezing point depression.
Freezing point depression is a colligative property of solutions, which means it depends on the number of solute particles, not their identity. It can be calculated using the formula:
ΔTf = Kf * m
Where:
ΔTf is the change in freezing point
Kf is the cryoscopic constant for the solvent
m is the molality of the solution
In this case, the freezing point of the pure benzene is 5.45°C, while the freezing point of the solution containing white phosphorus is 4.89°C. The change in freezing point (ΔTf) can be calculated as:
ΔTf = 5.45°C - 4.89°C = 0.56°C
The molality (m) of the solution can be calculated as:
m = moles of solute / mass of solvent in kg
To find the moles of solute, we need to know the molar mass of white phosphorus. The molar mass of white phosphorus is 124 g/mol.
moles of solute = mass of solute / molar mass of solute
= 0.675 g / 124 g/mol
≈ 0.00544 mol
mass of solvent in kg = 50.0 g / 1000 g/kg
= 0.050 kg
m = 0.00544 mol / 0.050 kg
= 0.1088 mol/kg
Now, we can substitute the values of Kf and m into the freezing point depression equation to solve for ΔTf:
0.56°C = 5.12 K mol⁻¹ kg * 0.1088 mol/kg
Simplifying the equation, we get:
0.56°C = 0.55616 K
Therefore, x = 4.
Hence, the value of x in the molecular formula (P)x is 4.
Freezing point of benzene = 5.45°C
Kf (benzene) = 5.12 K mol⁻¹ kg
Freezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89°C
To determine the molecular formula of white phosphorus in benzene, we can use the concept of freezing point depression.
Freezing point depression is a colligative property of solutions, which means it depends on the number of solute particles, not their identity. It can be calculated using the formula:
ΔTf = Kf * m
Where:
ΔTf is the change in freezing point
Kf is the cryoscopic constant for the solvent
m is the molality of the solution
In this case, the freezing point of the pure benzene is 5.45°C, while the freezing point of the solution containing white phosphorus is 4.89°C. The change in freezing point (ΔTf) can be calculated as:
ΔTf = 5.45°C - 4.89°C = 0.56°C
The molality (m) of the solution can be calculated as:
m = moles of solute / mass of solvent in kg
To find the moles of solute, we need to know the molar mass of white phosphorus. The molar mass of white phosphorus is 124 g/mol.
moles of solute = mass of solute / molar mass of solute
= 0.675 g / 124 g/mol
≈ 0.00544 mol
mass of solvent in kg = 50.0 g / 1000 g/kg
= 0.050 kg
m = 0.00544 mol / 0.050 kg
= 0.1088 mol/kg
Now, we can substitute the values of Kf and m into the freezing point depression equation to solve for ΔTf:
0.56°C = 5.12 K mol⁻¹ kg * 0.1088 mol/kg
Simplifying the equation, we get:
0.56°C = 0.55616 K
Therefore, x = 4.
Hence, the value of x in the molecular formula (P)x is 4.
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Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer?
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Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer?.
Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer?.
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Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer?, a detailed solution for Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer? has been provided alongside types of Molecular formula of white phosphorus in benzene is (P)x. What is the value of x based on following experiment?Freezing point of benzene = 5.45° C Kf (benzene) = 5.12 K mol-1 kgFreezing point of solution containing 0.675 g of white phosphorus in 50.0 g of benzene = 4.89° C.Correct answer is '4'. Can you explain this answer? theory, EduRev gives you an
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