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The vapour pressure of pure water at 25ºC is 23.76 torr. The vapour pressure of a solution containing 5.40 g of a nonvolatile substance in 90.0 g water is 23.32 torr. Compute the molecular weight of the solute.
    Correct answer is '57.4'. Can you explain this answer?
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    The vapour pressure of pure water at 25C is 23.76 torr. The vapour pre...


    nsolute = 0.094
    Msolute = 57.4
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    The vapour pressure of pure water at 25C is 23.76 torr. The vapour pre...
    Calculation of Molecular Weight of Solute

    Given:

    Vapour pressure of pure water at 25°C = 23.76 torr
    Vapour pressure of solution = 23.32 torr
    Mass of solute = 5.40 g
    Mass of solvent (water) = 90.0 g

    Step 1: Calculation of mole fraction of water

    Mole fraction of water = Number of moles of water / Total number of moles

    Number of moles of water = Mass of water / Molar mass of water
    = 90.0 g / 18.015 g/mol
    = 4.995 mol

    Number of moles of solute = Mass of solute / Molar mass of solute

    Step 2: Calculation of mole fraction of solute

    Mole fraction of solute = Number of moles of solute / Total number of moles

    Total number of moles = Number of moles of water + Number of moles of solute
    = 4.995 + (5.40 / Molecular weight of solute)

    Substituting the values in the equation for Raoult's Law:

    Ptotal = P°A × XA + P°B × XB

    Where Ptotal is the vapour pressure of the solution, P°A is the vapour pressure of pure solvent, XA is the mole fraction of solvent, P°B is the vapour pressure of the solute, and XB is the mole fraction of solute.

    Step 3: Calculation of molecular weight of solute

    23.32 = (23.76 x 4.995) + (P°B x (5.40 / Molecular weight of solute))

    P°B = 23.32 - (23.76 x 4.995)
    = 11.97 torr

    Substituting the values in the equation for the vapour pressure of the solute:

    P°B = XBMolal × Kb

    Where XBMolal is the molality of the solution and Kb is the molal boiling point elevation constant of water.

    Kb for water = 0.512 °C/m

    Molality of solution = Moles of solute / Mass of solvent (in kg)

    = (5.40 / Molecular weight of solute) / 0.090 kg
    = 0.6 / Molecular weight of solute

    Substituting the values in the equation for P°B:

    11.97 = (0.6 / Molecular weight of solute) x 0.512

    Molecular weight of solute = (0.6 / 11.97) x 0.512
    = 57.4

    Therefore, the molecular weight of the solute is 57.4.
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    Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..If VA ml of solution A having normality NA react just completely with VB ml of solution B having normalityNB thenEquation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.Mass of a substance expressed in gram equal to its equivalent man is gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.Q.Which of the following solutions, when mixed with 100 ml of 0.05 M NaOH, will give a neutral solution?

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    The vapour pressure of pure water at 25C is 23.76 torr. The vapour pressure of a solution containing 5.40 g of a nonvolatile substance in 90.0 g water is 23.32 torr. Compute the molecular weight of the solute.Correct answer is '57.4'. Can you explain this answer?
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    The vapour pressure of pure water at 25C is 23.76 torr. The vapour pressure of a solution containing 5.40 g of a nonvolatile substance in 90.0 g water is 23.32 torr. Compute the molecular weight of the solute.Correct answer is '57.4'. 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 The vapour pressure of pure water at 25C is 23.76 torr. The vapour pressure of a solution containing 5.40 g of a nonvolatile substance in 90.0 g water is 23.32 torr. Compute the molecular weight of the solute.Correct answer is '57.4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of pure water at 25C is 23.76 torr. The vapour pressure of a solution containing 5.40 g of a nonvolatile substance in 90.0 g water is 23.32 torr. Compute the molecular weight of the solute.Correct answer is '57.4'. Can you explain this answer?.
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