To calculate the weight/weight percentage (w/w %) of the acetic acid solution given its molarity and assuming the density is that of pure water (1000 g/L), follow these steps:
1. Determine the molar mass of acetic acid (CH₃COOH):
- Carbon (C): There are 2 carbons.
[tex]\[
2 \times 12.01 \, \text{g/mol} = 24.02 \, \text{g/mol}
\][/tex]
- Hydrogen (H): There are 4 hydrogens.
[tex]\[
4 \times 1.008 \, \text{g/mol} = 4.032 \, \text{g/mol}
\][/tex]
- Oxygen (O): There are 2 oxygens.
[tex]\[
2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol}
\][/tex]
- Sum of the molar mass:
[tex]\[
24.02 \, \text{g/mol} + 4.032 \, \text{g/mol} + 32.00 \, \text{g/mol} = 90.036 \, \text{g/mol}
\][/tex]
2. Calculate the volume and mass of the solution:
- Assume we have 1 liter of solution (since the molarity is given in moles per liter).
- Density of pure water is 1000 g/L.
- Mass of the solution:
[tex]\[
1 \, \text{L} \times 1000 \, \text{g/L} = 1000 \, \text{g}
\][/tex]
3. Determine the moles of acetic acid in the solution:
- Given molarity is 0.709 M (moles per liter):
[tex]\[
0.709 \, \text{moles/L} \times 1 \, \text{L} = 0.709 \, \text{moles}
\][/tex]
4. Convert the moles of acetic acid to grams:
- Using the molar mass of acetic acid:
[tex]\[
0.709 \, \text{moles} \times 90.036 \, \text{g/mol} = 63.835524 \, \text{g}
\][/tex]
5. Calculate the weight/weight percentage (w/w %) of the acetic acid in the solution:
- The mass of acetic acid in the solution is 63.835524 g.
- The total mass of the solution is 1000 g.
- The w/w % is:
[tex]\[
\left(\frac{63.835524 \, \text{g}}{1000 \, \text{g}}\right) \times 100 = 6.3835524
\][/tex]
Thus, the weight/weight percentage (w/w %) of the acetic acid in the solution is 6.3835524.
