You have a solution with 2.52 moles of isopropanol (C3H8O). The solution weighs 521 grams. What is the percent composition of isopropanol in the mixture?

The Percent composition of isopropanol in the mixture is 29.07 %

Explanation:

Step 1: Data given

Number of moles isopropanol (C3H8O) = 2.52 moles

Mass of the solution = 521 grams

Molar mass of isopropanol (C3H8O) = 60.1 g/mol

Step 2: Calculate mass of isopropanol

Mass isopropanol = moles isopropanol * molar mass isopropanol

Mass isopropanol = 2.52 moles * 60.1 g/mol

Mass isopropanol = 151.45 grams

Step 3: Calculate the percent composition of isopropanol in the mixture

Percent composition of isopropanol = (mass isopropanol / total mass of mixture) * 100 %

Percent composition of isopropanol = (151.45 grams / 521 grams ) * 100 %

Percent composition of isopropanol = 29.07 %

The Percent composition of isopropanol in the mixture is 29.07 %

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