Question

Please help me. I am unable to find help. Please please help me. I need help with question 3, not question 2

Answer 1

Question 2:

Answer for trial one: 8.0416 x 10⁻⁴ moles of OH⁻

Answer for trial two: 8.064 x 10⁻⁴ moles of OH⁻

Question 2:

The molar mass of the unknown acid for trial one = 249.45 g/mol

The molar mass of the unknown acid for trial two = 250.99 g/mol

Step-by-step explanation

Question 2:

Moles of OH⁻ for trial one = Molarity x Volume of OH in L for trial 2

Molarity = 0.112 M = 0.112 mol/L

Volume of OH in L for trial two = 7.18mL/1000 = 0.00718 L

∴ Moles of OH⁻ for trial one = 0.112 mol/L x 0.00718 L = 8.0416 x 10⁻⁴ mol

Also,

Moles of OH⁻ for trial two = Molarity x Volume of OH in L for trial 2

Molarity = 0.112 M = 0.112 mol/L

Volume of OH in L for trial 2 = 7.20mL/1000 = 0.0072 L

∴ Moles of OH⁻ for trial two = 0.112 mol/L x 0.0072 L = 8.064 x 10⁻⁴ mol

Question 3:

Note: You have been hinted to assume the number of moles for the weak acid to be equal to the number of moles of the base.

From question number 2;

Answer for trial one: 8.0416 x 10⁻⁴ moles of OH⁻

Answer for trial two: 8.064 x 10⁻⁴ moles of OH⁻

Therefore;

The number of moles of the weak acid for trial one will be assumed to be 8.0416 x 10⁻⁴ moles.

Also, the number of moles of the weak acid for trial two will be assumed to be 8.064 x 10⁻⁴ moles.

Also, from the given table;

Mass of the unknown acid for trial one = 0.2006 g

Mass of the unknown acid for trial two = 0.2024 g

Hence, the molar mass of the unknown acid for each trial can be calculated using the given formula:

For trial one:

`\begin{gathered} Molar\text{ }mass\text{ }of\text{ }the\text{ }unkown\text{ }acid\text{ }for\text{ }trial\text{ }one=\frac{0.2006\text{ }g}{8.0416*10⁻⁴\text{ }mol} \\ \\ Molar\text{ }mass\text{ }of\text{ }the\text{ }unkown\text{ }acid\text{ }for\text{ }trial\text{ }one=249.45\text{ }g\text{ /}mol \end{gathered}`

For trial two:

`\begin{gathered} Molar\text{ }mass\text{ }of\text{ }the\text{ }unkown\text{ }acid\text{ }for\text{ }trial\text{ }two=\frac{0.2024g}{8.064*10⁻⁴\text{ }mol} \\ \\ Molar\text{ }mass\text{ }of\text{ }the\text{ }unkown\text{ }acid\text{ }for\text{ }trial\text{ }two=250.99\text{ }g\text{/}mol \end{gathered}`