Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.
Amount of acetaminophen initially taken = 350 mg
Amount of acetaminophen obtained after recovery =185 mg
= 
= 52.9%
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
Answer:
aqueous acid is used as a reagent.
Explanation:
Addition of Grignard reagent in aldehyde and followed by the acidification give rise to the primary or secondary alcohol. when the formaldehyde is used than the primary alcohol is formed otherwise secondary alcohol is formed.
in this reaction we also use the aqueous acid for the acidification as a reagent. We add aqueous acid when ethanol is present. This is because ethanol is get converted in the presence of aqueous acid into the chloroethane.
First calculate the mole fraction of each substance:
Acetone: 2.88 mol ÷ (2.88 mol + 1.45 mol) = 0.665
Cyclohexane: 1.45 ÷ (2.88 mol + 1.45 mol) = 0.335
Raoult's Law: P(total) = P(acetone) · χ(acetone) + P(cyclohexane) · χ(cyclohexane).
P(total) = 229.5 torr · 0.665 + 97.6 torr · 0.335
P(total) = 185.3 torr
χ for acetone: 229.5 torr · 0.665 ÷ 185.3 torr = 0.823
χ for cyclohexane: 97.6 torr · 0.335 ÷ 185.3 torr = 0.177
