Respuesta :

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.

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