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A solid cube of edge length = 25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water. The boiling point of this solution is found to be 100.52°C (assume 100% ionisation of ionic compound). If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer) ?
(kb of water = 0.52 K kg mole-1, Avogadro number, NA = 6 x 1023 = 4.22)
    Correct answer is '100'. Can you explain this answer?
    Verified Answer
    A solid cube of edge length = 25.32 mm of an ionic compound which has ...
    Effective molality of solution = 1
    Hence, no. of moles of ionic solid in given cube = 0.5
    so, no. of formula units in given cube  
    no. of unit cells 
    no. of unit cells alongone edge of cube 
    If edge length of unit cell = 600 pm
    for NaCI type unit cell, a = 2 (r+ + r_)
    So r+ = 100 pm.
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    A solid cube of edge length = 25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water. The boiling point of this solution is found to be 100.52°C (assume 100% ionisation of ionic compound). If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer) ?(kb of water = 0.52 K kg mole-1, Avogadro number, NA = 6 x 1023,= 4.22)Correct answer is '100'. Can you explain this answer?
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    A solid cube of edge length = 25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water. The boiling point of this solution is found to be 100.52°C (assume 100% ionisation of ionic compound). If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer) ?(kb of water = 0.52 K kg mole-1, Avogadro number, NA = 6 x 1023,= 4.22)Correct answer is '100'. 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 A solid cube of edge length = 25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water. The boiling point of this solution is found to be 100.52°C (assume 100% ionisation of ionic compound). If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer) ?(kb of water = 0.52 K kg mole-1, Avogadro number, NA = 6 x 1023,= 4.22)Correct answer is '100'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid cube of edge length = 25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water. The boiling point of this solution is found to be 100.52°C (assume 100% ionisation of ionic compound). If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer) ?(kb of water = 0.52 K kg mole-1, Avogadro number, NA = 6 x 1023,= 4.22)Correct answer is '100'. Can you explain this answer?.
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