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

What is the amount of solute in 445 mL of 0.660 M KBr solution?

The amount of solute present in 445 mL of 0.660 M KBr solution is 34.951 grams.

Explanation:

Solute and Solution: The solute is the substance that is dissolved in the solvent to form a solution. In this case, the solute is KBr (Potassium Bromide). Molecular Weight: The molecular weight of KBr is 119.002 grams per mole. Molarity Calculation: Molarity (M) is defined as the number of moles of solute per liter of solution. It is calculated using the formula M = n/V, where n represents the number of moles and V represents the volume in liters. In this case, the molarity of the KBr solution is 0.660 M, which means there are 0.660 moles of KBr present in 1 liter of the solution. Calculating the amount of KBr: 0.660 mol KBr x 119.002 g KBr/1 mol 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 find the amount of KBr in 445 mL (0.445 L) of the solution. Therefore, 34.951 grams of KBr is the amount present in 445 mL of 0.660 M KBr solution.
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