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0.002 molal aqueous solution of an ionic compound [Co(NH3)5(N02)]CI freezes at - 0.00732° C. [Kf (H20) = 1.86° mol-1kg]. How many moles of ions does 1.0 mole of the salt produce on being dissolved in water?
    Correct answer is '2'. Can you explain this answer?
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    0.002 molal aqueous solution of an ionic compound [Co(NH3)5(N02)]CI fr...
     ΔTf = 0 - ( - 0.00732° ) = 0.00732°
    ΔTf = Molality x Kf x i
    0.00732 = 0.002 x 1.86 x i
    i = 1.97 = 2,0
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    0.002 molal aqueous solution of an ionic compound [Co(NH3)5(N02)]CI fr...
    Solution:

    Molality of the solution,
    m = 0.002 mol/kg

    Given,
    Kf (H2O) = 1.86 mol-1kg
    ΔTf = - 0.00732°C

    Using the formula,
    ΔTf = Kf × m

    - 0.00732 = 1.86 × 0.002
    - 0.00732 = 0.00372
    - 0.00732/0.00372 = 1.9677

    The value obtained is the van't Hoff factor (i) which is the number of particles into which one molecule of the solute dissociates in the solution.

    Therefore,
    i = 1.9677

    The value of i must be a whole number.
    So, the nearest whole number to 1.9677 is 2.

    Hence, the number of particles into which one molecule of the solute dissociates in the solution is 2.

    Therefore, one mole of the salt produces 2 moles of ions on being dissolved in water.
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    Read the passage given below and answer the following questions:All chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. All electrochemical cell reactions are also based on mole concept. For example, a 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrode. The amount of products formed can be calculated by using mole concept.The following questions are multiple choice questions. Choose the most appropriate answer:Q. In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at cathode while with Hg cathode it forms sodium amalgam

    0.002 molal aqueous solution of an ionic compound [Co(NH3)5(N02)]CI freezes at - 0.00732° C. [Kf (H20) = 1.86° mol-1kg]. How many moles of ions does 1.0 mole of the salt produce on being dissolved in water?Correct answer is '2'. Can you explain this answer?
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    0.002 molal aqueous solution of an ionic compound [Co(NH3)5(N02)]CI freezes at - 0.00732° C. [Kf (H20) = 1.86° mol-1kg]. How many moles of ions does 1.0 mole of the salt produce on being dissolved in water?Correct answer is '2'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 0.002 molal aqueous solution of an ionic compound [Co(NH3)5(N02)]CI freezes at - 0.00732° C. [Kf (H20) = 1.86° mol-1kg]. How many moles of ions does 1.0 mole of the salt produce on being dissolved in water?Correct answer is '2'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.002 molal aqueous solution of an ionic compound [Co(NH3)5(N02)]CI freezes at - 0.00732° C. [Kf (H20) = 1.86° mol-1kg]. How many moles of ions does 1.0 mole of the salt produce on being dissolved in water?Correct answer is '2'. Can you explain this answer?.
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