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A common recipe is to make 3% (w/vol) HCI in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCI is 12 moles per 1000 mL, how many mL of HCI need to be added to achieve a final volume of 150 mL of acid alcohol solution? Report your answer to two decimal places.

A common recipe is to make 3% (w/vol) HCI in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCI is 12 moles per 1000 mL, how many mL of HCI need to be added to achieve a final volume of 150 mL of acid alcohol solution? Report your answer to two decimal places.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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QUESTION 4 Microbiologists use acid-alcohol when performing a cell stain of the pathogen Mycobacterium tuberculosis. A common recipe is to make 3% (w/vol) HCI in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCI is 12 moles per 1000 mL, how many mL of HCI need to be added to achieve a final volume of 150 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. m₂ Mol HCI m₁ = 1000 mL 36.46 g HCI mol HCI V1 Then substitute this value of m, into the equation m₁v₁ = M₂V₂ and solve for v₁ = X x 100 %HCI M₂V2 m₁ 36.46 g HCI (3%) (v₂) m₂ Mol HCI 1000 mL (100) 1 mol HCI Solve for v₁ in a single step using Excel to avoid causing rounding error! = Save Answer
Transcribed Image Text:QUESTION 4 Microbiologists use acid-alcohol when performing a cell stain of the pathogen Mycobacterium tuberculosis. A common recipe is to make 3% (w/vol) HCI in ethanol. HCI has a formula weight of 36.46 grams per mole. If the stock solution of HCI is 12 moles per 1000 mL, how many mL of HCI need to be added to achieve a final volume of 150 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. m₂ Mol HCI m₁ = 1000 mL 36.46 g HCI mol HCI V1 Then substitute this value of m, into the equation m₁v₁ = M₂V₂ and solve for v₁ = X x 100 %HCI M₂V2 m₁ 36.46 g HCI (3%) (v₂) m₂ Mol HCI 1000 mL (100) 1 mol HCI Solve for v₁ in a single step using Excel to avoid causing rounding error! = Save Answer
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