Question

If you have 0.50 grams of p-aminophenol, how many mLs of water will be needed if you are adding 20 molar equivalents? How many mLs acetic anhydride will be required if you are adding 1.5 molar equivalents?

Answer 1

Answer: The volume of water required is 1.65 mL and the volume of acetic anhydride required is 0.65 mL

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(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:

`\text{Moles of p-aminophenol}=(0.50g)/(109.14g/mol)=4.58* 10^(-3)mol`

To calculate volume of solution, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}` ......(2)

• For water:

It is given that 20 molar equivalent so p-aminophenol is added:

Moles of water =
`(20* \text{Moles of p-aminophenol})=20* 4.58* 10^(-3)=9.16* 10^(-2)mol`

Mass of water is calculated by using equation 1, we get:

Molar mass of water = 18 g/mol

Mass of water =
`(9.16* 10^(-2)mol* 18g/mol)=1.65g`

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:

`1g/mL=\frac{1.65g}{\text{Volume of water}}\\\\\text{Volume of water}=(1.65g)/(1g/mL)=1.65mL`

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 =
`(1.5* \text{Moles of p-aminophenol})=1.5* 4.58* 10^(-3)=6.87* 10^(-3)mol`

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 =
`(6.87* 10^(-3)mol* 102.1g/mol)=0.701g`

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:

`1.08g/mL=\frac{0.701g}{\text{Volume of acetic anhydride}}\\\\\text{Volume of acetic anhydride}=(0.701g)/(1.08g/mL)=0.65mL`

Hence, the volume of acetic anhydride required is 0.65 mL