How would you prepare 1.30 L of a 0.30-M solution of each of the following? a HS
ID: 581496 • Letter: H
Question
How would you prepare 1.30 L of a 0.30-M solution of each of the following? a HSO, from "concentrated" (18 M) sulfturic acid Dilute[ ] mL ofconcentrated H2SO4 Submit HCI fron "concentrated" (12 M) reagent Dilute mL of concentrated HCI NICI, from the salt NiCIa 6H,0 DissolvegNiC 6H20 in water, and add water until the total volume of he HNO, from "concentrated (16 M) nitric acid DilutemL of concentrated HNO, eSodium carbonate from the pure solid Dissolveg Na CO, in water, and add water until the total volume of theExplanation / Answer
a)
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 18 M
V1 = initial volume = ?
C2 = final concentration = 0.30M
V2 = final volume = 1.30 L
C1V1 =C2V2 Thus
V1 = C2V2 /C1
Substitute the value in above equation
V1 = 0.30X 1.30/18 = 0.0216 L = 21.6 ml
take 21.6 ml 18 M H2SO4 solution and dilute it to 1.30 liter we get 0.30 M soltuion of H2SO4
b)
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 12 M
V1 = initial volume = ?
C2 = final concentration = 0.30M
V2 = final volume = 1.30 L
C1V1 =C2V2 Thus
V1 = C2V2 /C1
Substitute the value in above equation
V1 = 0.30X 1.30/12 = 0.0325 L = 32.5 ml
take 32.5 ml 12 M HCl solution and dilute it to 1.30 liter we get 0.30 M soltuion of HCl
c) no. of mole = molarity X volume of soluiton in liter
no. of mole NiCl2. 6H2O required to prepare 1.30 L 0.30 M solution = 0.30 X 1.30 = 0.39 mole
molar mas of NiCl2. 6H2O = 237.69 gm/mole then 0.39 mole of NiCl2. 6H2O = 0.39 X 237.69 = 92.70 gm
take 92.70 gm NiCl2. 6H2O and dilute upto 1.30 liter we get 1.30 liter 0. 30 M NiCl2. 6H2O solution
d)
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 16 M
V1 = initial volume = ?
C2 = final concentration = 0.30M
V2 = final volume = 1.30 L
C1V1 =C2V2 Thus
V1 = C2V2 /C1
Substitute the value in above equation
V1 = 0.30X 1.30/16 = 0.0244 L = 24.4 ml
take 24.4 ml 16 M HNO3 solution and dilute it to 1.30 liter we get 0.30 M soltuion of HNO3
e) no. of mole = molarity X volume of soluiton in liter
no. of mole Na2CO3 required to prepare 1.30 L 0.30 M solution = 0.30 X 1.30 = 0.39 mole
molar mas of Na2CO3 = 105.99 gm/mole then 0.39 mole of NiCl2. 6H2O = 0.39 X 105.99 = 41.33 gm
take 41.33 gm Na2CO3 and dilute upto 1.30 liter we get 1.30 liter 0. 30 M Na2CO3 solution
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