Prepare the following solutions by determiing the proper mass of solute (or volu
ID: 947575 • Letter: P
Question
Prepare the following solutions by determiing the proper mass of solute (or volume in the case where you are working from a stock solution) that will need to be transferred to the volumetric flask. Use diemensional analysis where apporpriate to do so. Describe how you would make each solution.
a. 150.0 mL of .50 M hydrochorlic acid solution from a concetrated stock solution of 12.0M hydrochloric acid.
b. 250.0 mL of a 1.50 M solution of potassium permanganate.
c. 2.00 L of a 3.0 Molar solution of perchloric acid from a 5.0M stock solution of perchloric acid.
d. 500.0 mL of a .300 molar solution of calcium hydroxide.
Explanation / Answer
part A
use the dilution formula
M1V1 = M2V2
we have a 12.0M solution we need to prepare 150 mL of 0.5M HCl solution
M1 = 12.0M, V1 = ?
M2 = 0.5M, V2 = 150 mL
12.0 x V2 = 0.5 x 150
V2 = 75/12 = 6.25 Ml
so take 6.25 ml of 12.0 M HCl trans fer in to 150 mL Volumetric solution make up up to 150 mark
so you need 150-6.25 = 143.75 ml of water
part B
we have to use the standard formula
molarity M = (weight / molar mass) x 1000 / volume in mL
M = 1.50, Weight = ?
molar mass of KMnO4 = 158.034 gram /mol
volume = 250 mL
put all these values in the above equation
1.50 = (weight/150.034) x 1000/250
weight = 1.50 x 150.34 /4
weight = 57.5025 grams
so weight the 57.5025 grams of KMnO4 trans fer in to 250 mL volumetric flask, dissolve in some amount of H2O the make up up to 250
Part C
this is also similar to Part A
we have a 5.0 M stock solution we need to prepare the 3.0M of 2L solution
5.0 x V1 = 3.0 x 2.0
V1 = 6/5
V1 = 1.2 Liters
take 1.2 liter of 5.0M per chloric acid transfer in to the 2L conical flask then dilute up to 2Liter mark
Part C
this is also simolar to part B
we have to use the standard formula
molarity M = (weight / molar mass) x 1000 / volume in mL
M = 0.3, Weight = ?
molar mass of Ca(OH)2 = 74.093 gram /mol
volume = 500 mL
put all these values in the above equation
0.3 = (weight/74.093) x 1000/500
weight = 0.3 x 79.093 /2
weight = 23.73 grams
so weight the 23.73 grams of Ca(OH)2 trans fer in to 250 mL volumetric flask, dissolve in some amount of H2O the make up up to 500 mL