A). You are asked to prepare a 500.0 mL of a 0.750 M iron (III) chloride solutio

Question

A). You are asked to prepare a 500.0 mL of a 0.750 M iron (III) chloride solution from a bottle that contains that solid. How many grams of iron (III) chloride would be needed to make this solution? Show work please.

B) = 0.016 L

C) A student obtains 25.0 mL of a 5.00M solution of HCl in a large beaker. The student then proceeds to add 400.0 mL of distilled water to the solution in the beaker. What is the concentration(molarity) of the diluted solution in the beaker?

Extra Question for maxiumum points

A solution is made by dissolving 3.00 grams of potassium nitrate (KNO3) in 500.0mL of water. Calculate the total concentration (Molarity) of ions in this solution

SHOW ALL WORK PLEASE

Explanation / Answer

A) moles = MV ( V in liters) = 0.75 x 500/100 = 0.375

hence FeCl3 mass required = moles x mol wt of FeCl3 = 0.375 x 162.2 = 60.825 gm

hence 60.825 gm FeCl3 is taken and transferred into measuring cylinder and it is filled with water upto 500 ml mark. Thus we get required sol

C) we use formula M1V1 = M2V2 for dilution purpose

( 5 x 25/1000) = M2 x ( 25 + 400) /1000

M2 = 0.294 Molar

D) moles of KNO3 = mass /mol mass = 3 / 101.1 = 0.0297

Molarity = ( 0.0297 x 1000/500) = 0.059

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