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 a…

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asked Oct 19, 2020 184k views

1 Answer

Giladbu · Answer 1 · 2020-10-23T12:24:14+0000

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)

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)

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

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