Asksia AI LOGO

Sia

青州's Question
Chemistry
Posted 6 months ago

Question: In the laboratory you dissolve 23.7 g23.7 \mathrm{~g} of chromium(III) acetate in a volumetric flask and add water to a total volume of 375 mL\mathrm{mL}.

In the laboratory you dissolve 23.7 g23.7 \mathrm{~g} of chromium(III) acetate in a volumetric flask and add water to a total volume of 375 mL375 \mathrm{~mL}.
Sign in to unlock the answer
Answer from Sia
Posted 6 months ago
Solution
1
Calculate the molar mass of chromium(III) acetate: The chemical formula for chromium(III) acetate is Cr(C2H3O2)3Cr(C_2H_3O_2)_3. The molar mass is calculated as follows: - Chromium (Cr): 1×51.9961g/mol1 \times 51.9961 \, \text{g/mol} - Carbon (C): 6×12.01g/mol6 \times 12.01 \, \text{g/mol} - Hydrogen (H): 9×1.008g/mol9 \times 1.008 \, \text{g/mol} - Oxygen (O): 6×16.00g/mol6 \times 16.00 \, \text{g/mol} Total molar mass = 51.9961+(6×12.01)+(9×1.008)+(6×16.00)=229.13g/mol51.9961 + (6 \times 12.01) + (9 \times 1.008) + (6 \times 16.00) = 229.13 \, \text{g/mol}
2
Determine the number of moles of chromium(III) acetate: Using the mass of chromium(III) acetate and its molar mass, we can find the number of moles: Number of moles=23.7g229.13g/mol=0.1034mol \text{Number of moles} = \frac{23.7 \, \text{g}}{229.13 \, \text{g/mol}} = 0.1034 \, \text{mol}
3
Calculate the molarity of the solution: Molarity (M) is defined as the number of moles of solute per liter of solution. The total volume of the solution is 375 mL, which is 0.375 L. Molarity=0.1034mol0.375L=0.276M \text{Molarity} = \frac{0.1034 \, \text{mol}}{0.375 \, \text{L}} = 0.276 \, \text{M}
Answer
The molarity of the chromium(III) acetate solution is 0.276M0.276 \, \text{M}
Key Concept
Molarity Calculation
Explanation
Molarity is calculated by dividing the number of moles of solute by the volume of the solution in liters. In this case, we first determined the molar mass of chromium(III) acetate, then used it to find the number of moles, and finally calculated the molarity.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Unlock Smarter Learning with AskSia Super!

Join Super, our all-in-one AI solution that can greatly improve your learning efficiency.

30% higher accuracy than GPT-4o
Entire learning journey support
The most student-friendly features
Study Other Question