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Stolb23 [73]
3 years ago
6

Calculate the Molarity (M) of an aqueous solution of HCl which is 35.38% HCl by mass and has a density of 1.161 g/mL.

Chemistry
1 answer:
kozerog [31]3 years ago
8 0

Answer:

11.2 M → [HCl]

Explanation:

Solution density = Solution mass / Solution volume

35.38 % by mass, is the same to say 35.38 g of solute in 100 g of solution.

Let's determine the moles of our solute, HCl

35.38 g . 1 mol/36.45 g = 0.970 moles

Let's replace the data in solution density formula

1.161 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.161 g/mL → 86.1 mL

Let's convert the volume to L → 86.1 mL . 1L / 1000 mL = 0.0861 L

Molarity (M) → mol/L = 0.970 mol / 0.0861 L → 11.2 M

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Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
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Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

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Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

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