Question

part 1 of 2
makenzie dissolved 22.0 g auf₃ in enough distilled water to make 425.00 ml of solution. perform the following calculations regarding this solution:
a. what is the molarity of this solution?

• this value will be used for subsequent calculations. remember to use the unrounded value of molarity in future calculations. enter units as \mol/l\.

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question 16
0/3 pts 5 19 details
a solution is prepared by dissolving 45 g of salt in 318 g of water. what is the mass percent of salt?
% m/m
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question 17
0/3 pts 5 19 details
calculate the mass percent of a sucrose solution that is made by mixing 6.25 grams of sucrose into water until the solution has a mass of 20 grams. give your answer as a whole number.

Explanation:

First, let's solve Question A (Molarity of \( \boldsymbol{AuF_3} \) solution):

Step1: Calculate moles of \( AuF_3 \)

Molar mass of \( Au \) is \( 196.97 \, \text{g/mol} \), \( F \) is \( 19.00 \, \text{g/mol} \). For \( AuF_3 \), molar mass \( M = 196.97 + 3\times19.00 = 196.97 + 57.00 = 253.97 \, \text{g/mol} \).
Moles \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{22.0 \, \text{g}}{253.97 \, \text{g/mol}} \approx 0.08662 \, \text{mol} \).

Step2: Convert volume to liters

Volume \( V = 425.00 \, \text{mL} = 0.42500 \, \text{L} \).

Step3: Calculate molarity

Molarity \( M = \frac{n}{V} = \frac{0.08662 \, \text{mol}}{0.42500 \, \text{L}} \approx 0.2038 \, \text{mol/L} \).

Step1: Find total mass of solution

Mass of salt \( m_{\text{salt}} = 45 \, \text{g} \), mass of water \( m_{\text{water}} = 318 \, \text{g} \).
Total mass \( m_{\text{solution}} = 45 + 318 = 363 \, \text{g} \).

Step2: Calculate mass percent

Mass percent \( = \frac{m_{\text{salt}}}{m_{\text{solution}}} \times 100 = \frac{45}{363} \times 100 \approx 12.40\% \).

Step1: Identify masses

Mass of sucrose \( m_{\text{sucrose}} = 6.25 \, \text{g} \), mass of solution \( m_{\text{solution}} = 20 \, \text{g} \).

Step2: Calculate mass percent

Mass percent \( = \frac{6.25}{20} \times 100 = 31.25\% \), rounded to whole number is \( 31\% \).

Answer:

\( 0.204 \, \text{mol/L} \) (or unrounded \( \approx 0.2038 \, \text{mol/L} \))

Now, Question 16 (Mass percent of salt):