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

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 $1.04 \mathrm{~g} / \mathrm{mL}$

Solution

Solution Steps

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

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

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

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

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

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

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

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

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

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 $\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2$ in one liter of solution:

\[ \text{Moles of solute} = \frac{74.88 \, \text{g}}{60.052 \, \text{g/mol}} = 1.247 \, \text{mol} \]

Thus, the molarity of the solution is:

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