Questions: What is the molarity of an aqueous solution that contains 0.0720 g C2H3O2 per gram of solution? The density of the solution is 1.04 g / mL

What is the molarity of an aqueous solution that contains 0.0720 g C2H3O2 per gram of solution? The density of the solution is 1.04 g / mL
Transcript text: What is the molarity of an aqueous solution that contains 0.0720 $\mathrm{g} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ per gram of solution ? The density of the solution is $1.04 \mathrm{~g} / \mathrm{mL}$
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Solution

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Solution Steps

Step 1: Calculate the mass of the solution in grams per liter

Given the density of the solution is 1.04g/mL1.04 \, \text{g/mL}, we can convert this to grams per liter:

1.04g/mL×1000mL/L=1040g/L 1.04 \, \text{g/mL} \times 1000 \, \text{mL/L} = 1040 \, \text{g/L}

Step 2: Calculate the mass of solute in grams per liter

The solution contains 0.0720g0.0720 \, \text{g} of C2H3O2\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2 per gram of solution. Therefore, the mass of the solute per liter of solution is:

0.0720g/g×1040g/L=74.88g/L 0.0720 \, \text{g/g} \times 1040 \, \text{g/L} = 74.88 \, \text{g/L}

Step 3: Calculate the molar mass of C2H3O2\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2

The molar mass of C2H3O2\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2 is calculated as follows:

2×12.01g/mol+3×1.008g/mol+2×16.00g/mol=60.052g/mol 2 \times 12.01 \, \text{g/mol} + 3 \times 1.008 \, \text{g/mol} + 2 \times 16.00 \, \text{g/mol} = 60.052 \, \text{g/mol}

Step 4: Calculate the molarity of the solution

Molarity (M) is defined as the number of moles of solute per liter of solution. We can calculate the number of moles of C2H3O2\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2 in one liter of solution:

Moles of solute=74.88g60.052g/mol=1.247mol \text{Moles of solute} = \frac{74.88 \, \text{g}}{60.052 \, \text{g/mol}} = 1.247 \, \text{mol}

Thus, the molarity of the solution is:

Molarity=1.247M \text{Molarity} = 1.247 \, \text{M}

Final Answer

Molarity=1.247M\boxed{\text{Molarity} = 1.247 \, \text{M}}

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