College

If 37.9 g of [tex]$C_2H_5OH$[/tex] (molar mass = 46.07 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the concentration of [tex]$C_2H_5OH$[/tex] in the resulting solution?

Answer :

To find the concentration of [tex]\( C_2H_5OH \)[/tex] (ethanol) in the solution, we can follow these steps:

1. Identify the known values:
- Mass of [tex]\( C_2H_5OH \)[/tex]: 37.9 g
- Molar mass of [tex]\( C_2H_5OH \)[/tex]: 46.07 g/mol
- Volume of the solution: 500.0 mL

2. Convert the volume of the solution from milliliters to liters:
[tex]\[
500.0 \, \text{mL} = 500.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.500 \, \text{L}
\][/tex]

3. Calculate the number of moles of [tex]\( C_2H_5OH \)[/tex]:
[tex]\[
\text{Moles of } C_2H_5OH = \frac{\text{Mass of } C_2H_5OH}{\text{Molar mass of } C_2H_5OH} = \frac{37.9 \, \text{g}}{46.07 \, \text{g/mol}} \approx 0.8227 \, \text{mol}
\][/tex]

4. Calculate the concentration of [tex]\( C_2H_5OH \)[/tex]:
[tex]\[
\text{Concentration} = \frac{\text{Moles of } C_2H_5OH}{\text{Volume of solution in L}} = \frac{0.8227 \, \text{mol}}{0.500 \, \text{L}} \approx 1.6453 \, \text{mol/L}
\][/tex]

So, the concentration of [tex]\( C_2H_5OH \)[/tex] in the solution is approximately [tex]\( 1.6453 \, \text{mol/L} \)[/tex].