Microbiologists use acid-alcohol when performing a cell stain of the pathogen Mycobacterium tuberculosis. A common recipe is to make 3%wewol) HCl in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCl is 12 moles per 1000 mL, how many mL of HCl need to be added to achieve a final volume of 246 mL of acid alcohol solution? Report your answer to two decimal places. Algebra can be used to solve this problem! In this context, the percentages act as concentrations. Therefore, it is possible to use m¡vj = m2v2 and use the percentages as the concentration values. We can solve for m, by converting the molarity of HCI into a percentage using its formula weight. 36.46 g HCI m1 Mol HCI х 100 %3D %НСІ = Tu mol HCI 1000 mL Then substitute this value of m, into the equation m2V2 and solve for v1 m2V2 (3%) (v2) V = 36.46 g HCI m1 Mol HCI 1 mol HCI m1 |(100) 1000 mL Solve for v, in a single step using Excel to avoid causing rounding error!

Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter15: Acid–base Equilibria
Section: Chapter Questions
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Solve for v1 in a single step using Excel to avoid causing rounding error!

 
Microbiologists use acid-alcohol when performing a cell stain of the pathogen
Mycobacterium tuberculosis.
A common recipe is to make 3%(w/vol) HCl in ethanol. HCI has a formula weight of 36.46
grams per mole. If the stock solution of HCl is 12 moles per 1000 mL, how many ml of
HCl need to be added to achieve a final volume of 246 mL of acid alcohol solution? Report
your answer to two decimal places.
Algebra can be used to solve this problem!
In this context, the percentages act as concentrations. Therefore, it is possible to use m1v1 = m2V2 and use the
percentages as the concentration values. We can solve for m, by converting the molarity of HCI into a percentage
using its formula weight.
36.46 g HCI
m1 Mol HCI
m1
х 100 %3D %HСІ
%3D
mol HCI
1000 mL
Then substitute this value of m1 into the equation
mįv1 = m2V2
and solve for v1
m2V2
(3%) (v2)
V1
m1
36.46 g HCI
m1 Mol HCI
(100)
1 mol HCI
1000 mL
Solve for v, in a single step using Excel to avoid causing rounding error!
Transcribed Image Text:Microbiologists use acid-alcohol when performing a cell stain of the pathogen Mycobacterium tuberculosis. A common recipe is to make 3%(w/vol) HCl in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCl is 12 moles per 1000 mL, how many ml of HCl need to be added to achieve a final volume of 246 mL of acid alcohol solution? Report your answer to two decimal places. Algebra can be used to solve this problem! In this context, the percentages act as concentrations. Therefore, it is possible to use m1v1 = m2V2 and use the percentages as the concentration values. We can solve for m, by converting the molarity of HCI into a percentage using its formula weight. 36.46 g HCI m1 Mol HCI m1 х 100 %3D %HСІ %3D mol HCI 1000 mL Then substitute this value of m1 into the equation mįv1 = m2V2 and solve for v1 m2V2 (3%) (v2) V1 m1 36.46 g HCI m1 Mol HCI (100) 1 mol HCI 1000 mL Solve for v, in a single step using Excel to avoid causing rounding error!
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