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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.

2 Answers

4 votes

Answer:


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.

User Reva
by
8.4k points
6 votes

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.

User Carolanne
by
7.7k points