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DerKrebs [107]
3 years ago
13

First, get about 20 mL of the quinine stock solution in a clean beaker to prepare 250 mL of about 2 ppm quinine in 0.05 M H2SO4.

Calculate the actual final concentration based on your dilution of the stock solution.
Chemistry
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

Volume of acid, Va=250mL; Volume of quinine,Vb=20mL; Molarity of acid, Ma=0.05M.

Molar mass of acid= H2+S+O4= 2+32+(16X4)= 2+32+64=98g

Concentration of acid, Ca= Molar mass of acid/ Ma =98/0.05=1960g/mol

Explanation: To calculate concentration of quinine, Cb is as follow

Va*Ca=Vb*Cb

∴ Cb=Va*Ca/Vb =250*1960/20 =24500g/mol

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<h3>Answer:</h3>

0.127 mol Au

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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<u>Step 1: Define</u>

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[Solve] moles Au

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 25.0 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
  2. [DA] Multiply/Divide [Cancel out units:                                                          \displaystyle 0.126923 \ mol \ Au

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.126923 mol Au ≈ 0.127 mol Au

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