The printed label on a bottle of commercial vinegar states that the acetic acid

Question

The printed label on a bottle of commercial vinegar states that the acetic acid
concentration is 5%.
(a) If the manufacturer had reported two significant figures in the concentration, what
would the percentage range be so that all values would round to 5%?
(b) Calculate the concentration in molarity of the upper and lower values from (a).
The molecular weight of acetic acid is 60.05 g/mol. How many digits can you
report in your answers?
(c) What assumption did you make to calculate the concentrations in part (b)?
(d) What is the percentage difference in the concentrations you calculated in part (b)?
Explain what value you chose for the denominator in this calculation?

PLEASE HELP!

Explanation / Answer

a)   5.0 t0 5.4 % range has two sig fig and would round off to 5%

(b)

For value 5.0% 5g acetic acid in 100 ml of solution

Moles of acetic acid = weight (grams) /molecular weight = 5/60.05 = 0.083 moles

Volume of solution = 100 ml = 100/1000 = 0.1 Litres [as 1000ml = 1L]

Molarity = moles/volume in litres = 0.083 /0.1 = 0.83 M (answer)

For value 5.4% 5.4g acetic acid in 100 ml of solution

Moles of acetic acid = weight (grams) /molecular weight = 5 .4/60.05 = 0.0899 moles

Volume of solution = 100 ml = 100/1000 = 0.1 Litres [as 1000ml = 1L]

Molarity = moles/volume in litres = 0.0899 /0.1 = 0.899 M (answer) = 0.9 M (answer)

c) We assumed that volume of solution is 100 ml

d) % difference in concentration =[( 0.9 -0.830 ) / 0.830]*100= (0.07 /0.830 ) *100 = 8.4%

e) We Chose 0.830 M ( concentration calculated using 5.0%) for denominator as that is the true or accepted value given on bottle

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.