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If you have 450mL of 1.25M solution of KBr, how much KBr does it contain?

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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:


\frac{450mL\text{ . 1.25moles}}{1000mL}=0.56\text{ moles}

- Second, with the molar mass of KBr (119g/mol), we find the grams of KBr:


\frac{0.56\text{mol . 119g}}{1\text{mol}}=66.64\text{ g}

- So, there are 66.64 g of KBr.

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