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A solution containing 30 g of non-volatile solute in exactly 90 g of water has a vapor pressure of 21.85 mm of Hg at 250 C. Further 18 g of water is added to the solution. The resulting solution has a vapor pressure of 22.15 mm of Hg at same temp. Calculate the molecular weight (g/mol) of solute: [rounded up to two decimal places]
    Correct answer is between '67,69'. Can you explain this answer?
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    A solution containing 30 g of non-volatile solute in exactly 90 g of w...




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    A solution containing 30 g of non-volatile solute in exactly 90 g of w...
    Solution:

    Given data:

    Mass of solute (m1) = 30 g
    Mass of solvent (m2) = 90 g
    Total mass of solution (m1 + m2) = 120 g
    Vapor pressure of solution (P1) = 21.85 mm Hg
    Mass of water added (m3) = 18 g
    Total mass of solution after adding water (m1 + m2 + m3) = 138 g
    Vapor pressure of resulting solution (P2) = 22.15 mm Hg
    Temperature (T) = 25°C = 298 K

    1. Calculate the mole fraction of solute in the original solution:

    Mole fraction of solute (X1) = m1 / (m1 + m2)
    X1 = 30 / (30 + 90) = 0.25

    2. Calculate the mole fraction of solvent in the original solution:

    Mole fraction of solvent (X2) = m2 / (m1 + m2)
    X2 = 90 / (30 + 90) = 0.75

    3. Calculate the vapor pressure of pure water at 25°C:

    The vapor pressure of pure water (P0) at 25°C = 23.76 mm Hg

    4. Calculate the vapor pressure of solvent in the original solution:

    P2/P0 = X2
    P2 = X2 * P0
    P2 = 0.75 * 23.76
    P2 = 17.82 mm Hg

    5. Calculate the vapor pressure of solute in the original solution:

    P1 = P0 * X2 * (1 + (m1 / m2) * (1 / Kf))
    where Kf is the freezing point depression constant of water (1.86 °C/m)

    Rearranging the above equation, we get:

    (m1 / m2) = [((P1 / P0) - 1) / (X2 * 1 / Kf)]
    (m1 / m2) = [((21.85 / 23.76) - 1) / (0.75 * 1.86)]
    (m1 / m2) = 0.109

    m1 = 0.109 * m2
    m1 = 0.109 * 90
    m1 = 9.81 g

    Molar mass of solute (M1) = (m1 / n1)

    n1 = (m1 / M1)
    M1 = (m1 / n1)
    M1 = (9.81 / 0.3048)
    M1 = 32.16 g/mol

    6. Calculate the mole fraction of solvent in the resulting solution:

    Mole fraction of solvent (X2) = (m2 / (m1 + m2 + m3))
    X2 = (90 / (30 + 90 + 18))
    X2 = 0.625

    7. Calculate the mole fraction of solute in the resulting solution:

    Mole fraction of solute (X1) = (m1 / (m1 + m2 + m3))
    X1 = (30 / (30 + 90 + 18))
    X1 = 0.192

    8. Calculate the vapor pressure of solvent in the resulting solution:

    P
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    A solution containing 30 g of non-volatile solute in exactly 90 g of w...
    Let moles of solute in both case = y
    for case 1:-

    moles of H2O = 5
    therefore
    (Pi-Pf)/Pi = y/(y+5)
    (Pi-21.85)/Pi = y/(y+5)
    by cross multiplication we have
    Piy+5Pi-21.85y-21.85*5=Piy

    5Pi=21.85y+21.85*5 ............ eq1

    for case 2:-

    moles of H2O = 6
    therefore
    (Pi-Pf)/Pi = y/(y+6)
    (Pi-22.15)/Pi = y/(y+6)
    by cross multiplication we have
    Piy+6Pi-22.15y-22.15*6=Piy

    6Pi=22.15y+22.15*6 ............ eq2

    now 6*eq1-5*eq2 we have

    30Pi-30Pi=y*(21.85*6-22.15*5)+30*(21.85-22.15)

    y=0.4422

    now mol. wt. = given wt. / y = 30/0.4422 = 67.83
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    A solution containing 30 g of non-volatile solute in exactly 90 g of water has a vapor pressure of 21.85 mm of Hg at 250 C. Further 18 g of water is added to the solution. The resulting solution has a vapor pressure of 22.15 mm of Hg at same temp. Calculate the molecular weight (g/mol) of solute: [rounded up to two decimal places]Correct answer is between '67,69'. Can you explain this answer?
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    A solution containing 30 g of non-volatile solute in exactly 90 g of water has a vapor pressure of 21.85 mm of Hg at 250 C. Further 18 g of water is added to the solution. The resulting solution has a vapor pressure of 22.15 mm of Hg at same temp. Calculate the molecular weight (g/mol) of solute: [rounded up to two decimal places]Correct answer is between '67,69'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A solution containing 30 g of non-volatile solute in exactly 90 g of water has a vapor pressure of 21.85 mm of Hg at 250 C. Further 18 g of water is added to the solution. The resulting solution has a vapor pressure of 22.15 mm of Hg at same temp. Calculate the molecular weight (g/mol) of solute: [rounded up to two decimal places]Correct answer is between '67,69'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution containing 30 g of non-volatile solute in exactly 90 g of water has a vapor pressure of 21.85 mm of Hg at 250 C. Further 18 g of water is added to the solution. The resulting solution has a vapor pressure of 22.15 mm of Hg at same temp. Calculate the molecular weight (g/mol) of solute: [rounded up to two decimal places]Correct answer is between '67,69'. Can you explain this answer?.
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