QUESTION 4Microbiologists use acid-alcohol when performing a cell stain of the pathogenMycobacterium tuberculosis.A common recipe is to make 3% (w/vol) HCI in ethanol. HCI has a formula weight of36.46 grams per mole. If the stock solution of HCI is 12 moles per 1000 mL, howmany mL of HCI need to be added to achieve a final volume of 150 mL of acidalcohol 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 touse m₁v₁ = m₂v₂ and use the percentages as the concentration values. We can solvefor m, by converting the molarity of HCI into a percentage using its formula weight.m₂ Mol HCIm₁ =1000 mL36.46 g HCImol HCIV1Then substitute this value of m, into the equationm₁v₁ = M₂V₂and solve for v₁=Xx 100 %HCIM₂V2m₁36.46 g HCI(3%) (v₂)m₂ Mol HCI1000 mL(100)1 mol HCISolve for v₁ in a single step using Excel to avoid causing rounding error!=Save Answer

Given:Initial concentration of HCl solution prepared = 3 %.V2 = 150 mL.Moles of HCl in stock…

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

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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