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LenaWriter [7]
3 years ago
9

A student obtained 1.69 g of pure caffeine following recrystallization from a crude sample, which originally weighed 2.51 g. Wha

t is the percent recovery of pure caffeine from crude?
Chemistry
1 answer:
Savatey [412]3 years ago
5 0

Answer:

The recovery of pure caffeine is 67.3% of the original, crude sample.

Explanation:

Step 1: Data given

Mass of pure caffeine = 1.69 grams

Original mass of sample = 2.51 grams

Step 2: What is the percent recovery of pure caffeine from crude?

Percentage = (mass of recovery / original mass) *100%

Percentage = (1.69 grams/2.51 grams) *100%

Percentage = 67.3 %

The recovery of pure caffeine is 67.3% of the original, crude sample.

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What is the pH of the solution formed when 25 mL of 0.173 M NaOH is added to 35 mL of 0.342 M HCl?
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Answer:

0.89

Explanation:

You are mixing an acid and a base, so there will be a neutralization reaction.

One of the reactants will be in excess, so we must determine its concentration and then calculate the pH.

<em>Moles of NaOH </em>

Moles of NaOH = 25 × 0.173

Moles of NaOH = 4.32 mmol

===============

<em>Moles of HCl </em>

Moles of HCl = 35 × 0.342

Moles of HCl = 12.0 mmol

===============

<em>Amount of excess reactant </em>

              NaOH + HCl ⟶ NaCl +H₂O

<em>n</em>/mmol:   4.32      12.0

The 4.32 mmol of NaOH reacts completely with 4.32 mmol of HCl.

Excess HCl = 12.0 – 4.32

Excess HCl = 7.6 mmol

===============

<em>Concentration of the excess HCl </em>

Total volume = 25 + 35

Total volume = 60 mL

<em>c </em>= millimoles HCl/millilitres HCl

<em>c</em> = 7.6/60

<em>c</em> = 0.13 mol/L

===============

<em>Calculate the pH </em>

The HCl dissociates completely to hydronium ions, so

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pH = -log[H₃O⁺]

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