how many ions and moles does 113 mL of .490 M aluminum chloride have?
Molarity = moles/liters so moles = molarity x liters. 113mL = .113L so .113 x .490 = 0.05537moles. as far as the ions go, i think you would have to look at a periodic table and convert the molecular weight or something. that's a strange question, im not sure if you are wording it right. hope this helps
AlCl3.
Wendell is correct for moles AlCl3 = M x L = ??
Then moles Al = the same as moles AlCl3 since there is 1 mole Al per 1 mole AlCl3.
For moles Cl, it is 3xmoles AlCl3 since there are 3 Cl atoms per molecule AlCl3.
To determine the number of ions and moles in a given solution of aluminum chloride, you need to know the chemical formula of aluminum chloride and use the provided concentration and volume.
The chemical formula for aluminum chloride is AlCl3, indicating that each molecule of aluminum chloride contains one aluminum ion (Al3+) and three chloride ions (Cl-).
First, let's calculate the number of moles of aluminum chloride in the given solution:
Step 1: Convert the volume from milliliters (mL) to liters (L):
113 mL ÷ 1000 = 0.113 L
Step 2: Use the formula for calculating moles:
moles = concentration (M) × volume (L)
moles = 0.490 M × 0.113 L = 0.05537 moles of AlCl3 (rounded to five decimal places)
Therefore, there are 0.05537 moles of aluminum chloride in 113 mL of a 0.490 M solution.
Next, let's determine the number of ions present in the solution:
Step 3: Multiply the number of moles of aluminum chloride by the number of ions produced per molecule of aluminum chloride.
For aluminum ions (Al3+):
0.05537 moles × 1 = 0.05537 moles of Al3+ (rounded to five decimal places)
For chloride ions (Cl-):
0.05537 moles × 3 = 0.16611 moles of Cl- (rounded to five decimal places)
Finally, let's convert the moles of ions to the number of ions:
Step 4: Multiply the number of moles by Avogadro's number (6.022 × 10^23 ions per mole) to get the number of ions:
For aluminum ions (Al3+):
0.05537 moles × 6.022 × 10^23 ions/mole = 3.3355 × 10^22 ions of Al3+ (rounded to four significant figures)
For chloride ions (Cl-):
0.16611 moles × 6.022 × 10^23 ions/mole = 9.9933 × 10^22 ions of Cl- (rounded to four significant figures)
Therefore, in 113 mL of a 0.490 M aluminum chloride solution, there are approximately 3.34 × 10^22 aluminum ions (Al3+) and 9.99 × 10^22 chloride ions (Cl-).