NTA Abhyas JEE Main2020ChemistrySolutionsPractice
Compound P d C l 4 . 6 H 2 O is a hydrated complex; 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex ( K f for water = 1.86 K kg m o l - 1 )
Options
- A[ P d ( H 2 O ) 6 ] C l 4
- B[ P d ( H 2 O ) 4 C l 2 ] C l 2 . 2 H 2 O
- C[ P d ( H 2 O ) 3 C l 3 ] C l . 3 H 2 O
- D[ P d ( H 2 O ) 2 C l 4 ] . 4 H 2 O
Correct answer
C. [ P d ( H 2 O ) 3 C l 3 ] C l . 3 H 2 O
Step-by-step solution
Δ T = i × K f × m ( 273 - 269.28 ) = i × 1.86 × 1 3.72 = i × 1.86 i = 2 α = i - 1 n - 1 1 = 2 - 1 n - 1 or n = 2 Thus, the complex should give two ions in the solution, i.e., the complex will be [ P d ( H 2 O ) 3 C l 3 ] C l . 3 H 2 O .