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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
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    Consider two liquids B and C that form an ideal solutlion. We hold the temperature fixed at some value T that is above the freezing points of B and C .We shall plot the systems pressure P and against XB the overall mole fraction of B in the system :Where nbland nbvare the number of moles of B in the liquid and vapor phases, respectively. For a close system xB, is fixed, although nBland nBvmay vary. Let the system be enclosed in a cylinder fitted with a piston and immersed in a constant-temperature bath.To see what the P-versus-xB phase diagram looks like, let us initially set the external pressure on the piston high enough for the system to be entirely liquid (point A in figur e) As the pressure is lowered below that at A, the system eventually reaches a pressure where the liquid just begins to vaporizes (point D). At point D, the liquid has composition xlbwhere xlbat D is equal to the overall mole fraction xB since only an infinitesimal amount of liquid has vapourized. What is the composition of the first vapour that comes off ? Raoults law, Pb=xvbPobrelates the vapour-phase mole fractions to the liquid composition as follows:Where Poband Pocare the vapour pressures of pure B and pure C at T, where the systems pressure P equals the sum PB + Pc of the partial pressures, where and the vapor is assumed ideal.Let B be the more volatile component, meaning that PobPocAbove equation then shows that Xvb/XvcXlb/XlcThe vapor above an ideal solution is richer than the liquid in the more volatile component. Equations (1) and (2) apply at any pressure where liquid -vapor equilibrium exists, not just at point D. Now let us isothermally lower the pressure below point D, causing more liquid to vaporize. Eventually, WE reach point F in figure, where the last drop of liquid vaporizes. Below F, we have only vapor. For any point or the line between D and F liquid and vapor phases coexist in equilibrium.Q.Two liquids A and B have the same molecular weight and form an ideal solution. The solution has a vapor pressure of 700 Torrs at 80C. It is distilled till 2/3rd of the solution (2/3rd moles out of total moles) is collected as condensate.The composition of the condensate is xA = 0.75 and that of the residue is XA= 0.30. If the vapor pressure of the residue at 80C is 600 Torrs, find the original composition of the liquid.

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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?
    Question Description
    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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