NTA Abhyas JEE Main2020ChemistrySome Basic Concepts of ChemistryPractice
1 g of a complex [Cr(H 2 O) 5 Cl]Cl 2 ⋅ H 2 O (mol. Wt. 266.5) was passed through a cation exchanger to produce HCl . The acid liberated was diluted to 1 litre. The normality of this acid solution is
Options
- A5 × 1 0 – 3 N
- B7.5 × 1 0 – 3 N
- C7.5 × 1 0 – 2 N
- D7.5 × 1 0 – 1 N
Correct answer
B. 7.5 × 1 0 – 3 N
Step-by-step solution
Mole of complex = W m o l . wt. = 1 266.5 1 mole of complex produces 2 mole HCl [Cr(H 2 O) 5 Cl]Cl 2 → Cr(H 2 O)Cl] 2+ 2Cl – Mole of HCl formed = 1 266.5 × 2 N HCl = n V × n − factor = 2 266 .5×1 = 7 .5 × 10 –3 N