Calculate the final concentration of each of the following diluted solutions. A)
ID: 229714 • Letter: C
Question
Calculate the final concentration of each of the following diluted solutions.
A) 4.0 L of a 4.0 M HNO3 solution is added to water so that the final volume is 8.0 L .
Express your answer using two significant figures.
B) Water is added to 0.45 L of a 6.0 M KOH solution to make 2.0 L of a diluted KOH solution.
Express your answer using two significant figures
C) A 80.0 mL sample of 8.0 % (m/v) NaOH is diluted with water so that the final volume is 200.0 mL.
Express your answer using two significant figures.
D) A 8.0 mL sample of 50.0 % (m/v) acetic acid (HC2H3O2) solution is added to water to give a final volume of 25 mL .
Express your answer using two significant figures.
Explanation / Answer
A. Initial Final
M1= 4.0M M2=
V1= 4.0L V2= 8.0L
for dilution M1V1 = M2V2
M2= 4.0x4.0/8.0 = 2.0M
final concentration of HNO3 = 2.0M
B) Initial Final
M1= 6.0M M2=
V1= 0.45L V2= 2.0L
M1V1= M2V2
M2= 6.0x0.45/2.0 = 1.35M
final concentration of KOH= 1.35M = 1.4M
C) 8.0%(m/v) NaOH solution means 8.0 grams of NaOH is dissolved in 100 ml of solution.
mass of NaOH= 8.0 grams
molar mass of NaOH= 40 gram/mol
volume = 100 ml
Molarity = mass/molar mass x1000/v in ml
Molarity = 8.0/40 x1000/100 = 2.0M
Initial concentration of NaOH= 2.0M
initial volume of NaOH= 80.0ml
final volume= 200.0ml
final concentration of NaOH= 2.0x80.0/200.0 = 0.8M
final concentration of NaOH= 0.80M
D 50.0%(m/V) acetic acid means 50.0 grams of acetic acid is present in 100 ml of solution
mass of acetic acid = 50.0 grams
molar mass of acetic acid= 60 gram/mole
volume= 100 ml
M= 50.0/60 x1000/100= 8.33M
inital concentration of acetic acid= 8.33M
initial volume= 8.0ml
final volume=25ml
final concentration of acetic acid= 8.33x8.0/25 = 2.666 M
final concentration of acetic acid = 2.7M
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