Question

Question #5f140

Answer 1

The stock solution was made using 47.1 g of potassium phosphate.

You can work backwards from the solution of known molarity to the stock solution by focusing on number of moles of solute.

You know that your final solution has a volume of 125 mL and a concentration of 0.1250 M, which means that it contains

`C = n/V => n = C * V`

`n = "0.1250 M" * 125 * 10^(-3)"L" = "0.1563 moles"` `K_3PO_4`

The final solution was obtained by diluting a 35.25-mL sample taken from the stock solution. This tells you that the 35.25-mL sample must contain an equal amount of moles of potassium phosphate.

When you dilute a solution, the volume changes from the addition of more solvent, but the number of moles of solute remains constant.

Therefore, you know that you have 0.1563 moles of potassium phosphate in the 35.25 mL sample.

Since this sample was taken directly from the stock solution, you can determine the number of moles of potassium phosphate present in the stock solution by using a simple ratio

`500cancel("mL") * ("0.1563 moles "K_3PO_4)/(35.25cancel("mL")) = "0.2217 moles"` `K_3PO_4`

Use potassium phosphate's molar mass to determine how many grams were needed to get this many moles

`0.2217cancel("moles "K_3PO_4) * "212.27 g"/(1cancel("mole "K_3PO_4)) = color(green)("47.1 g")`

SIDE NOTE Once again, I left the answer with three sig figs, despite the fact that you only gave one sig fig for the volume of the stock solution.

Judging by the number of sig figs given for the other values, I'd say that you should have had something like 500. mL, or maybe 500.0 mL.