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Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)
(Kb = 2.16 K kg mol-1)
    Correct answer is '158'. Can you explain this answer?
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    Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Applications of colligative properties are very useful in day to day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.Given: Freezing point depression constant of water (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol-1Molecular weight of ethanol = 46 g mol-1In answering the following question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate. Correct answer is '376.2'. Can you explain this answer?

    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?

    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.100 ml of 0.1 M H3 P02 is titrated with 0.1 M NaOH. The volume of NaOH needed will be

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    Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. Can you explain this answer?
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    Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. 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 Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. Can you explain this answer?.
    Solutions for Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
    Here you can find the meaning of Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. Can you explain this answer?, a detailed solution for Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. Can you explain this answer? has been provided alongside types of Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of pure ether. The molecular weight of the substance (in g mol-1) will be (Nearest integer)(Kb = 2.16 K kg mol-1)Correct answer is '158'. Can you explain this answer? tests, examples and also practice JEE tests.
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