Question

A chemist must dilute 53.6mL of 5.23 M aqueous potassium iodide (KI) solution until the concentration falls to 2.00 M He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits.

Answer 1

`0.139 L`

Step-by-step explanation:

To dilute the potassium iodide solution

--->
`M_1V_1=M_2V_2`

where
`M_1` is the initial concentration of the solution,
`V_1` is the initial volume of the solution,
`M_2` is the final concentration of the solution, and
`V_2` is the final volume of the solution.

We can rearrange this formula to solve for
`V_2`:

`V_2 = (M_1V_1)/(M_2)`

Substituting the given values, we get:

`V_2=(5.23 *53.6)/(2.00)`

• Note that we need to convert the initial volume from milliliters to liters to ensure consistent units in the calculation.

53.6 mL is equal to 0.0536 L.

`V_2=(5.23*0.0536)/(2.00)`

`V_2 = 0.139`

Therefore, the final volume of the solution is 0.139 L.

Answer 2

Answer: 0.0866

Step-by-step explanation:

We know that there are
`(5.23)(53.6/1000)=0.280328` moles of KI.

If we want the molarity to be 2.00 M, then there must be
`(0.280328)/(2.00)=0.140164` L of solution.

There are initially 0.0536 L of solution, so another
`0.140164-0.0536=0.0866` mL of solution must be added.