There are 66.64 g of KBr.
- First, we need to calculate the moles of KBr that are contained in 450 mL of 1.25 M solution of KBr:
[tex]\frac{450mL\text{ . 1.25moles}}{1000mL}=0.56\text{ moles}[/tex]- Second, with the molar mass of KBr (119g/mol), we find the grams of KBr:
[tex]\frac{0.56\text{mol . 119g}}{1\text{mol}}=66.64\text{ g}[/tex]- So, there are 66.64 g of KBr.
