Answer:
We have to weigh 52.8 g of BaCl₂·2H₂O, add it to a 2.00 L flask and add water until reaching the final volume.
Step-by-step explanation:
Describe the preparation of 2.00 L of 0.108 M BaCl₂ from BaCl₂·2H₂O. (244.3 g/mol).
Step 1: Calculate the moles of BaCl₂
We need to prepare 2.00 L of a solution that contains 0.108 moles of BaCl₂ per liter of solution.
2.00 L × 0.108 mol/L = 0.216 mol
Step 2: Calculate the moles of BaCl₂·2H₂O that contain 0.216 moles of BaCl₂
The molar ratio of BaCl₂·2H₂O to BaCl₂ is 1:1. The moles of BaCl₂·2H₂O required are 1/1 × 0.216 mol = 0.216 mol.
Step 3: Calculate the mass corresponding to 0.216 mol of BaCl₂·2H₂O
The molar mass of BaCl₂·2H₂O is 244.3 g/mol.
0.216 mol × 244.3 g/mol = 52.8 g
We have to weigh 52.8 g of BaCl₂·2H₂O, add it to a 2.00 L flask and add water until reaching the final volume.