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The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL
of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to
    Correct answer is '6'. Can you explain this answer?
    Most Upvoted Answer
    The volume (in mL) of 0.1 M AgNO3 required for complete precipitation ...
    Number of ionisable Cl- in [Cr(H2O)5Cl]Cl2 is 2
    Millimoles of Cl- = 30 x 0.01 ´ 2 = 0.6
    Millimoles of Ag+ required = 0.6
    0.6 = 0.1 V
    V = 6 ml
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    Community Answer
    The volume (in mL) of 0.1 M AgNO3 required for complete precipitation ...
    One mole of the given complex dissociates into two chloride ions.


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    The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mLof 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close toCorrect answer is '6'. Can you explain this answer?
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    The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mLof 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close toCorrect answer is '6'. 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 volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mLof 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close toCorrect answer is '6'. 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 volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mLof 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close toCorrect answer is '6'. Can you explain this answer?.
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