Question

Question was provided via image upload.

Explanation:

Calculate the average volume

Using the Average Volume Calculation knowledge point

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$$ V_{\text{avg}} = \frac{2.07\text{ mL}}{1000\text{ mL/L}} = 0.00207\text{ L} $$

Identify missing information for concentration

To solve Question 2, we need the mole ratio from "part 4" and the concentration of the standard solution (compound #1). Since these values are not visible in the cropped image, we can express the concentration of the unknown acid symbolically:
Let \(M_{\text{std}}\) be the molarity of the standard solution.
Let \(r\) be the stoichiometric ratio from part 4:

$$ r = \frac{\text{moles of unknown acid}}{\text{moles of standard solution}} $$

The volume of the unknown acid is \(V_{\text{acid}} = 10.0\text{ mL} = 0.0100\text{ L}\).

Express unknown acid concentration

The moles of standard solution used are:

$$ n_{\text{std}} = M_{\text{std}} \times 0.00207\text{ L} $$

The moles of unknown acid are:

$$ n_{\text{acid}} = r \times n_{\text{std}} = r \times M_{\text{std}} \times 0.00207\text{ L} $$

The molarity of the unknown acid \(M_{\text{acid}}\) is:

$$ M_{\text{acid}} = \frac{n_{\text{acid}}}{0.0100\text{ L}} = 0.207 \times r \times M_{\text{std}} $$

Express mass of unknown acid

To solve Question 3, we need the molar mass (\(MW\)) of the unknown acid:

$$ \text{moles of acid in sample} = n_{\text{acid}} = M_{\text{acid}} \times 0.0100\text{ L} $$

$$ \text{grams of acid} = n_{\text{acid}} \times MW $$

Answer:

Question 1

The average volume of standard solution added is:

• Average volume in mL: \(2.07\text{ mL}\)
• Average volume in L: \(0.00207\text{ L}\)

Question 2

Due to the cropped image, the concentration of the standard solution and the reaction ratio from part 4 are missing. Using the average volume of \(0.00207\text{ L}\), the formula to calculate the concentration is:

$$ M_{\text{acid}} = \frac{M_{\text{standard}} \times 0.00207\text{ L} \times \text{mole ratio}}{0.0100\text{ L}} $$

Question 3

The mass of the unknown acid in the sample is calculated using:

$$ \text{grams} = M_{\text{acid}} \times 0.0100\text{ L} \times \text{Molar Mass of acid (g/mol)} $$