Answer:
98.68%
Step-by-step explanation:
Step 1: Write the balanced equation
KBr + AgNO₃ ⇒ AgBr + KNO₃
Step 2: Calculate the moles corresponding to 814.5 mg (0.8145 g) of AgBr
The molar mass of AgBr is 187.77 g/mol.
0.8145 g × 1 mol/187.77 g = 4.338 × 10⁻³ mol
Step 3: Calculate the moles of KBr needed to produce 4.338 × 10⁻³ moles of AgBr
The molar ratio of KBr to AgBr is 1:1. The moles of KBr needed are 1/1 × 4.338 × 10⁻³ mol = 4.338 × 10⁻³ mol.
Step 4: Calculate the pure mass corresponding to 4.338 × 10⁻³ moles of KBr
The molar mass of KBr is 119.00 g/mol.
4.338 × 10⁻³ mol × 119.00 g/mol = 0.5162 g
Step 5: Calculate the purity of KBr
0.5162 g of KBr are in a 0.5231 g-sample. The purity of KBr is:
P = 0.5162 g/0.5231 g × 100% = 98.68%