Respuesta :
Answer: The volume of water required is 1.65 mL and the volume of acetic anhydride required is 0.65 mL
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For p-aminophenol:
Given mass of p-aminophenol = 0.50 g
Molar mass of p-aminophenol = 109.14 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of p-aminophenol}=\frac{0.50g}{109.14g/mol}=4.58\times 10^{-3}mol[/tex]
To calculate volume of solution, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex] ......(2)
- For water:
It is given that 20 molar equivalent so p-aminophenol is added:
Moles of water = [tex](20\times \text{Moles of p-aminophenol})=20\times 4.58\times 10^{-3}=9.16\times 10^{-2}mol[/tex]
Mass of water is calculated by using equation 1, we get:
Molar mass of water = 18 g/mol
Mass of water = [tex](9.16\times 10^{-2}mol\times 18g/mol)=1.65g[/tex]
Now, calculating the volume of water by using equation 2, we get:
Density of water = 1 g/mL
Mass of water = 1.65 g
Putting values in equation 2, we get:
[tex]1g/mL=\frac{1.65g}{\text{Volume of water}}\\\\\text{Volume of water}=\frac{1.65g}{1g/mL}=1.65mL[/tex]
Hence, the volume of water required is 1.65 mL
- For acetic anhydride:
It is given that 1.5 molar equivalent so p-aminophenol is added:
Moles of acetic anhydride = [tex](1.5\times \text{Moles of p-aminophenol})=1.5\times 4.58\times 10^{-3}=6.87\times 10^{-3}mol[/tex]
Mass of acetic anhydride is calculated by using equation 1, we get:
Molar mass of acetic anhydride = 102.1 g/mol
Mass of acetic anhydride = [tex](6.87\times 10^{-3}mol\times 102.1g/mol)=0.701g[/tex]
Now, calculating the volume of acetic anhydride by using equation 2, we get:
Density of acetic anhydride = 1.08 g/mL
Mass of acetic anhydride = 0.701 g
Putting values in equation 2, we get:
[tex]1.08g/mL=\frac{0.701g}{\text{Volume of acetic anhydride}}\\\\\text{Volume of acetic anhydride}=\frac{0.701g}{1.08g/mL}=0.65mL[/tex]
Hence, the volume of acetic anhydride required is 0.65 mL
