How many mL of a 0.124 moles/L solution of potassium permanganate contain 0.722 g of the salt?

Question

Wyatt Adkins · Accepted Answer

#"36.8 mL"#

As you are aware, molarity is the quantity of solute in moles per liter of a particular solution.

In your case, a solution of sodium permanganate, #"KMnO"_4#, is said to have a molarity of #"0.124 mol L"^(-1)#. This tells you that every #"1 L"# of this solution will contain #0.124# moles of solute.

To solve this problem, one interesting approach would be to change the solution's molarity from moles per liter to grams per milliliter.

Potassium permanganate has a molar mass of #"158.034 g mol"^(-1)#, which means that #0.124# moles of potassium permanganate would be equivalent to

#0.124 color(red)(cancel(color(black)("moles KMnO"_4))) * "158.034 g"/(1color(red)(cancel(color(black)("mole KMnO"_4)))) = "19.60 g"#

Additionally, you are aware that you have

#color(blue)(ul(color(black)("1 L" = 10^3"mL")))#

which indicates that the solution's molarity corresponds to a concentration of

#("0.124 moles KMnO"_4)/("1 L solution") = ("19.60 g KMnO"_4)/(10^3"mL solution")#

Now you're ready to use this as a conversion factor to determine how many milliliters of solution would contain #"0.722 g"# of potassium permanganate

#0.722 color(red)(cancel(color(black)("g KMnO"_4))) * (10^3"mL solution")/(19.60color(red)(cancel(color(black)("g KMnO"_4)))) = color(darkgreen)(ul(color(black)("36.8 mL solution")))#

Three sig figs are used to round the result.

Jeremiah Adams · Answer

undefined To find the volume of the solution, we first need to calculate the number of moles of potassium permanganate (KMnO4) in 0.722 g using its molar mass. Then, we can use the molarity of the solution to find the volume.

1. Calculate the number of moles of KMnO4:
   $ \text{Molar mass of KMnO}_4 = 158.034 \, \text{g/mol} $
   $ \text{Number of moles} = \frac{0.722 \, \text{g}}{158.034 \, \text{g/mol}} $

2. Use the formula for molarity:
   $ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume (in liters)}} $
   We rearrange the formula to find the volume:
   $ \text{Volume (in liters)} = \frac{\text{Number of moles}}{\text{Molarity}} $

3. Convert the volume from liters to milliliters.

Performing the calculations:

1. $ \text{Number of moles} = \frac{0.722 \, \text{g}}{158.034 \, \text{g/mol}} \approx 0.00457 \, \text{moles} $

2. $ \text{Volume (in liters)} = \frac{0.00457 \, \text{moles}}{0.124 \, \text{moles/L}} \approx 0.0369 \, \text{L} $

3. $ \text{Volume (in milliliters)} = 0.0369 \, \text{L} \times 1000 = 36.9 \, \text{mL} $

Therefore, 36.9 mL of the 0.124 moles/L solution of potassium permanganate contain 0.722 g of the salt.