how much chloride ions does 10.00 mL of .500M BaCl2 have?
To determine the number of chloride ions in 10.00 mL of a 0.500 M BaCl2 solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (L)
First, let's convert the volume from milliliters (mL) to liters (L):
10.00 mL = 10.00 mL * (1 L / 1000 mL) = 0.01000 L
Now we can calculate the moles of BaCl2 using the molarity equation:
0.500 M = moles / 0.01000 L
Rearranging the equation to solve for moles:
moles = 0.500 M * 0.01000 L
moles = 0.00500 moles
BaCl2 dissociates into one barium ion (Ba2+) and two chloride ions (2Cl-). Since the stoichiometry of the compound is 1:2, we can multiply the moles of BaCl2 by 2 to find the moles of chloride ions:
moles of chloride ions = 0.00500 moles * 2 = 0.01000 moles
Finally, we can convert moles back to the number of chloride ions by using Avogadro's number, which is 6.022 x 10^23 ions per mole:
number of chloride ions = moles * Avogadro's number
number of chloride ions = 0.01000 moles * (6.022 x 10^23 ions/mole)
number of chloride ions = 6.022 x 10^21 chloride ions
Therefore, 10.00 mL of a 0.500 M BaCl2 solution contains approximately 6.022 x 10^21 chloride ions.