Question

How many mmoles of each ion are present in 448 mL of a 0.0840 M solution of K2Cr2O7?

Answer 1

To determine the number of mmoles of each ion present in the solution, multiply the molarity by the volume of the solution. For K2Cr2O7, there are 2 K+ ions and 1 Cr2O7^2- ion per mole of the compound.

Step-by-step explanation:

To determine the number of mmoles of each ion present, we need to consider the formula of K2Cr2O7. This compound dissociates into three ions when it dissolves in water: 2 K+ ions and 1 Cr2O7^2- ion. Since we have a 0.0840 M solution of K2Cr2O7, the concentration of each ion will be the same.

Therefore, the number of mmoles of each ion present can be calculated using the molarity and volume of the solution. First, convert the volume of the solution from mL to L by dividing by 1000: 448 mL = 0.448 L.

Now, multiply the molarity (0.0840 M) by the volume (0.448 L) to find the number of moles of K2Cr2O7. Since we have 2 K+ ions for every 1 mole of K2Cr2O7, we can determine the number of mmoles of K+ ions. Similarly, since we have 1 Cr2O7^2- ion for every 1 mole of K2Cr2O7, we can determine the number of mmoles of Cr2O7^2- ions.

Answer 2

mmoles of potassium ion = 75.264 mmoles

mmoles of dichromate ion = 37.632 mmoles

Step-by-step explanation:

Considering:-

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Moles =Molarity * {Volume\ of\ the\ solution}`

For
`K_2Cr_2O_7` :

Molarity = 0.0840 M

Volume = 448 mL

The mmoles of
`K_2Cr_2O_7` are;-

`Moles =0.0840 * 448\ mmoles=37.632\ mmoles`

In 1 mmole of
`K_2Cr_2O_7`, 2 mmoles of potassium ion and 1 mmole of dichromate ion is present.

In 37.632 mmole of
`K_2Cr_2O_7`, 2*37.632 mmoles of potassium ion and 1*37.632 mmole of dichromate ion is present.

mmoles of potassium ion = 75.264 mmoles

mmoles of dichromate ion = 37.632 mmoles