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Grade: 10
        
Please help me with the question . I m in urgent need of it 
one year ago

Answers : (1)

Arun
23008 Points
							
6.40 mL AgNO3 = 0.00640 L AgNO3 
0.00640 L (0.200 M) = 0.00640 L (0.200 mol/L) = 0.00128 mol AgNO3 

For every mole of AgNO3 that dissolves in solution, there is one mole of Ag+ formed, so you have 0.00128 mol Ag+. 

For every mole of Ag+ precipitated as AgBr, one mole of Br- is precipitated, so you have 0.00128 mol Br-. 

0.00128 mol Br- (79.9 g/mol) = 0.102 g Br- in the compound. 

To find the percentage by mass: 
0.102 g Br- ÷ 2.00 g total = 0.0511 = 5.11% bromide in the sample.
one year ago
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