Questions: What is the concentration (M) of CH3OH in a solution prepared by dissolving 5.9 g of CH3OH in sufficient water to give exactly 230 mL of solution?

What is the concentration (M) of CH3OH in a solution prepared by dissolving 5.9 g of CH3OH in sufficient water to give exactly 230 mL of solution?

Transcript text: What is the concentration (M) of $\mathrm{CH}_{3} \mathrm{OH}$ in a solution prepared by dissolving 5.9 g of $\mathrm{CH}_{3} \mathrm{OH}$ in sufficient water to give exactly 230 mL of solution? 1.59 0.00159 0.800 11.9 0.00119

Solution

Solution Steps

Step 1: Calculate the Molar Mass of CH$_3$OH

The chemical formula for methanol is CH$_3$OH. To find its molar mass, we sum the atomic masses of its constituent atoms:

Carbon (C): 1 atom × 12.01 g/mol = 12.01 g/mol
Hydrogen (H): 4 atoms × 1.008 g/mol = 4.032 g/mol
Oxygen (O): 1 atom × 16.00 g/mol = 16.00 g/mol

Adding these together gives the molar mass of CH$_3$OH: \[ 12.01 + 4.032 + 16.00 = 32.042 \, \text{g/mol} \]

Step 2: Calculate the Number of Moles of CH$_3$OH

To find the number of moles of CH$_3$OH, use the formula: \[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \] Substituting the given values: \[ \text{moles of CH}_3\text{OH} = \frac{5.9 \, \text{g}}{32.042 \, \text{g/mol}} \approx 0.1841 \, \text{mol} \]

Step 3: Calculate the Concentration of CH$_3$OH

Concentration (Molarity, M) is defined as the number of moles of solute per liter of solution. The volume of the solution is given as 230 mL, which is equivalent to 0.230 L. Therefore, the concentration is: \[ \text{Concentration (M)} = \frac{\text{moles of CH}_3\text{OH}}{\text{volume of solution (L)}} \] Substituting the values: \[ \text{Concentration (M)} = \frac{0.1841 \, \text{mol}}{0.230 \, \text{L}} \approx 0.8004 \, \text{M} \]