Ethylene dibromide (C2H4Br2) and 1,2-dibromopropane (C3H6Br2) form a series of ideal solutions over the whole range of compositions and at 85°C, the vapour pressures of these two pure liquids are 173 torr and 127 torr, respectively. If 10.0 g of ethylene bromide is dissolved in 80.0 g of 1,2-dibromopropane, then mole fraction of ethylene dibromide in the vapour phase is K × 10-1. Find the value of K. (Nearest integer)Correct answer is '2'. Can you explain this answer?

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Ethylene dibromide (C₂H₄Br₂) and 1,2-dibromopropane (C₃H₆Br₂) form a series of ideal solutions over the whole range of compositions and at 85°C, the vapour pressures of these two pure liquids are 173 torr and 127 torr, respectively. If 10.0 g of ethylene bromide is dissolved in 80.0 g of 1,2-dibromopropane, then mole fraction of ethylene dibromide in the vapour phase is K × 10^-1. Find the value of K. (Nearest integer)

Correct answer is '2'. Can you explain this answer?

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Ethylene dibromide (C2H4Br2) and 1,2-dibromopropane (C3H6Br2) form a ...

Solution:

Given:
- Ethylene dibromide (C2H4Br2) and 1,2-dibromopropane (C3H6Br2) form a series of ideal solutions.
- At 85°C, the vapour pressures of ethylene dibromide and 1,2-dibromopropane are 173 torr and 127 torr, respectively.
- 10.0 g of ethylene dibromide is dissolved in 80.0 g of 1,2-dibromopropane.

To find:
The mole fraction of ethylene dibromide in the vapor phase (K × 10-1).

Explanation:

Step 1: Calculate the moles of ethylene dibromide and 1,2-dibromopropane:

Given that the molecular weight of ethylene dibromide (C2H4Br2) is:
2 * atomic weight of carbon (C) + 4 * atomic weight of hydrogen (H) + 2 * atomic weight of bromine (Br)
= (2 * 12.01 g/mol) + (4 * 1.01 g/mol) + (2 * 79.90 g/mol)
= 187.93 g/mol

The moles of ethylene dibromide can be calculated using the formula:
moles = mass / molecular weight
moles of ethylene dibromide = 10.0 g / 187.93 g/mol

Given that the molecular weight of 1,2-dibromopropane (C3H6Br2) is:
3 * atomic weight of carbon (C) + 6 * atomic weight of hydrogen (H) + 2 * atomic weight of bromine (Br)
= (3 * 12.01 g/mol) + (6 * 1.01 g/mol) + (2 * 79.90 g/mol)
= 201.94 g/mol

The moles of 1,2-dibromopropane can be calculated using the formula:
moles = mass / molecular weight
moles of 1,2-dibromopropane = 80.0 g / 201.94 g/mol

Step 2: Calculate the mole fraction of ethylene dibromide:

The total moles of the solution can be calculated by adding the moles of ethylene dibromide and 1,2-dibromopropane.

Total moles = moles of ethylene dibromide + moles of 1,2-dibromopropane

Now, calculate the mole fraction of ethylene dibromide using the formula:
Mole fraction = moles of ethylene dibromide / total moles

Step 3: Calculate the partial pressure of ethylene dibromide in the vapor phase:

The vapor pressure of a component in an ideal solution is directly proportional to its mole fraction.

Partial pressure of ethylene dibromide = Mole fraction of ethylene dibromide * Vapor pressure of ethylene dibromide

Step 4: Calculate the value of K:

The mole fraction of ethylene dibromide in the vapor phase is given as K × 10-1.

So, we need to convert the mole fraction

Answered by Infinity Academy · View profile

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Ethylene dibromide (C2H4Br2) and 1,2-dibromopropane (C3H6Br2) form a series of ideal solutions over the whole range of compositions and at 85°C, the vapour pressures of these two pure liquids are 173 torr and 127 torr, respectively. If 10.0 g of ethylene bromide is dissolved in 80.0 g of 1,2-dibromopropane, then mole fraction of ethylene dibromide in the vapour phase is K × 10-1. Find the value of K. (Nearest integer)Correct answer is '2'. Can you explain this answer?

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Ethylene dibromide (C2H4Br2) and 1,2-dibromopropane (C3H6Br2) form a series of ideal solutions over the whole range of compositions and at 85°C, the vapour pressures of these two pure liquids are 173 torr and 127 torr, respectively. If 10.0 g of ethylene bromide is dissolved in 80.0 g of 1,2-dibromopropane, then mole fraction of ethylene dibromide in the vapour phase is K × 10-1. Find the value of K. (Nearest integer)Correct answer is '2'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Ethylene dibromide (C2H4Br2) and 1,2-dibromopropane (C3H6Br2) form a series of ideal solutions over the whole range of compositions and at 85°C, the vapour pressures of these two pure liquids are 173 torr and 127 torr, respectively. If 10.0 g of ethylene bromide is dissolved in 80.0 g of 1,2-dibromopropane, then mole fraction of ethylene dibromide in the vapour phase is K × 10-1. Find the value of K. (Nearest integer)Correct answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ethylene dibromide (C2H4Br2) and 1,2-dibromopropane (C3H6Br2) form a series of ideal solutions over the whole range of compositions and at 85°C, the vapour pressures of these two pure liquids are 173 torr and 127 torr, respectively. If 10.0 g of ethylene bromide is dissolved in 80.0 g of 1,2-dibromopropane, then mole fraction of ethylene dibromide in the vapour phase is K × 10-1. Find the value of K. (Nearest integer)Correct answer is '2'. Can you explain this answer?.

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