How many chloride ions are present in 65.5 mL of 0.210 M AlCl3 solution
Well, let's see. AlCl3 is a compound made up of one aluminum ion (Al3+) and three chloride ions (Cl-). The formula tells us that for every one AlCl3 molecule, there are three chloride ions.
Now, to find out how many chloride ions are present in 65.5 mL of 0.210 M AlCl3 solution, we need to use the formula:
Number of Cl- ions = Concentration of AlCl3 × Volume of AlCl3 × Number of Cl- ions per AlCl3 molecule
Plugging in the numbers, we get:
Number of Cl- ions = 0.210 M × 65.5 mL × 3 Cl- ions per AlCl3 molecule
Well, well, well... my calculator seems to have disappeared! Oh, wait, here it is in my pocket. Let me just do some quick calculations...
*rustling sounds*
*tapping noises*
Okay, so the number of chloride ions present in 65.5 mL of 0.210 M AlCl3 solution is approximately equal to... drumroll, please...
**6.50625 × 10^21 chloride ions!**
Wow, that's a whole lot of chloride ions. They sure like to hang out in that AlCl3 solution!
To determine the number of chloride ions present in the given solution, we need to use the concentration and volume of the solution.
First, we need to convert the volume of the solution from milliliters to liters:
65.5 mL = 65.5 / 1000 = 0.0655 L
Next, we can use the concentration of AlCl3 to calculate the chloride ion concentration. Since there are three chloride ions for every one AlCl3 molecule, the chloride ion concentration will be three times the concentration of AlCl3:
Chloride ion concentration = (0.210 M) x 3 = 0.630 M
Now, we can use the chloride ion concentration and the volume of the solution to find the number of chloride ions present. The formula to calculate the number of moles is:
moles = concentration (M) x volume (L)
moles of chloride ions = (0.630 M) x (0.0655 L) = 0.04132 moles
Finally, to find the number of chloride ions, we use Avogadro's number, which states that there are approximately 6.022 x 10^23 particles (atoms, molecules, or ions) in one mole. Therefore:
Number of chloride ions = moles of chloride ions x Avogadro's number
Number of chloride ions = 0.04132 moles x 6.022 x 10^23 = 2.49 x 10^22 chloride ions
Therefore, there are approximately 2.49 x 10^22 chloride ions present in 65.5 mL of 0.210 M AlCl3 solution.
To find the number of chloride ions present in a given volume and concentration of AlCl3 solution, we need to use the concept of molarity and stoichiometry.
Molarity (M) is defined as the number of moles of solute (in this case, AlCl3) per liter of solution. Therefore, we need to convert the given volume of the solution from milliliters (mL) to liters (L) and then calculate the number of moles of AlCl3.
First, let's convert the volume from mL to L:
65.5 mL = 65.5/1000 L = 0.0655 L
Next, we can use the formula for molarity (M) to calculate the number of moles of AlCl3:
Molarity (M) = Moles of solute / Volume of solution (in L)
Rearranging the formula, we get:
Moles of solute = Molarity (M) × Volume of solution (in L)
Now, let's calculate the number of moles of AlCl3:
Number of moles of AlCl3 = 0.210 M × 0.0655 L = 0.013755 mol
Next, we need to consider the stoichiometry of AlCl3. According to the balanced chemical equation for AlCl3, for every 1 mole of AlCl3, there are 3 moles of chloride ions (Cl-).
So, if we have 0.013755 moles of AlCl3, the number of moles of chloride ions would be:
Number of moles of Cl- = 3 × 0.013755 mol = 0.041265 mol
Finally, we can convert the number of moles of chloride ions to the number of chloride ions using Avogadro's number.
Number of chloride ions = (0.041265 mol) × (6.022 × 10^23 ions/mol) = 2.4866 × 10^22 chloride ions
Therefore, there are approximately 2.49 × 10^22 chloride ions present in 65.5 mL of a 0.210 M AlCl3 solution.
mols AlCl3 = M x L = ?
mols Cl^- = 3x that
There are 6.02E23 Cl^- in 1 mol Cl^-.