JEE Main 2020 — Solutions Question with Solution
From: JEE Main 2020 (Online) 6th September Morning Slot
Question
The elevation of boiling point of 0.10 m
aqueous CrCl3.xNH3 solution is two times that
of 0.05 m aqueous CaCl2 solution. The value of
x is ______.
[Assume 100% ionisation of the complex and CaCl2, coordination number of Cr as 6, and that all NH3 molecules are present inside the coordination sphere]
[Assume 100% ionisation of the complex and CaCl2, coordination number of Cr as 6, and that all NH3 molecules are present inside the coordination sphere]
Enter your answer
Show full solutionCorrect answer: 5
Correct answer
5
Step-by-step explanation
Molality of CaCl2 solution = 0.05 m
Tb = i Kb m = 3 × Kb × 0.05 = 0.15 Kb
Molality of CrCl3.xNH3 = 0.10 m
Tb' = i Kb 0.10
Given, Tb' = 2Tb
i Kb 0.10 = 0.15 Kb
i = 3
Co-ordination number of Cr is 6.
[Cr(NH3 )x.Cl6-x]Cl3-6+x [Cr(NH3 )xCl] + (3-6+x)Cl–
i = 3 = 1 + 3 - 6 + x
x = 5
Tb = i Kb m = 3 × Kb × 0.05 = 0.15 Kb
Molality of CrCl3.xNH3 = 0.10 m
Tb' = i Kb 0.10
Given, Tb' = 2Tb
i Kb 0.10 = 0.15 Kb
i = 3
Co-ordination number of Cr is 6.
[Cr(NH3 )x.Cl6-x]Cl3-6+x [Cr(NH3 )xCl] + (3-6+x)Cl–
i = 3 = 1 + 3 - 6 + x
x = 5
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