A student determined the average molarity of their sample of acetic acid to be 0.709 M. Calculate the (w/w)% of the solution. Assume the density of this solution is that of pure water, [tex]d_{H_2O} = 1000 \, \text{g/L}[/tex]. Enter a number without the '%' symbol.



Answer :

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.