number of Cu2+ ions in 48 L of 3.8 M copper(II) chloride
moles Cu = M x L.
1 mole contains 6.022 x 10^23 atoms Cu^+2
To determine the number of Cu2+ ions in 48 L of 3.8 M copper(II) chloride, we need to use the formula:
Number of moles = Volume (in L) x Concentration (in M)
First, we calculate the number of moles of copper(II) chloride (CuCl2) in the solution:
Number of moles of CuCl2 = Volume (in L) x Concentration (in M)
= 48 L x 3.8 M
= 182.4 moles
Since there is one Cu2+ ion per CuCl2 molecule, the number of Cu2+ ions will be the same as the number of moles of CuCl2. Therefore, there are 182.4 Cu2+ ions in 48 L of 3.8 M copper(II) chloride.
To determine the number of Cu2+ ions in the given solution, we need to follow a few steps:
Step 1: Find the number of moles of Cu2+ ions.
To do this, we can use the given molarity (M) and volume (L) of the solution.
Number of moles of Cu2+ ions = Molarity × Volume
Given: Molarity = 3.8 M
Volume = 48 L
Substituting these values into the equation, we get:
Number of moles of Cu2+ ions = 3.8 M × 48 L
Step 2: Convert moles to the number of Cu2+ ions.
Since copper(II) chloride dissociates into one Cu2+ ion and two Cl- ions, the number of Cu2+ ions will be equal to the number of moles of Cu2+ ions.
Number of Cu2+ ions = Number of moles of Cu2+ ions
Step 3: Calculate the number of Cu2+ ions using Avogadro's number.
Avogadro's number states that there are 6.022 × 10^23 entities (in this case, ions) in one mole of a substance.
Number of Cu2+ ions = Number of moles of Cu2+ ions × Avogadro's number
Substituting the values, we get:
Number of Cu2+ ions = (3.8 M × 48 L) × 6.022 × 10^23
Calculating this expression will give you the final answer.