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How many grams of solute are present in 445 mL of 0.660 M KBr?

User Adjoa
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1 Answer

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22 votes

Answer: 34.951 g of KBr is the amount in 445mL of .660M KBr.

Explanation: The solute is what is dissolved in the solvent. In this case the solute is KBr. The Molecular weight of KBr is 119.002g

Molarity = n/V, so .660 M = x moles/1L, so it's .660 moles/L.

.660 mol KBr x 119.002 g KBr/1 mole KBr = 78.541 grams of KBr in one mole of KBr.

78.541 g KBr/1 mol KBr = X g of KBr/0.445 L KBr cross multiply and divide to get grams of KBr in 445mL (which is the same as 0.445 L

Therefore, 34.951 g of KBr is the amount in 445mL of .660M KBr.

User Siim Nelis
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