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Lilit [14]
4 years ago
14

What is the molarity of a stock solution if 60 mL were used to make 150 mL of a .5M solution

Chemistry
1 answer:
kirza4 [7]4 years ago
6 0

Answer:

The answer to your question is 1.25 M

Explanation:

Data

Molarity 1 = ?

Volume 1 = 60 ml

Molarity 2 = 0.5 M

Volume 2 = 150 ml

Process

1.- Write the dilution formula

       Molarity 1 x Volume 1 = Molarity 2 x Volume 2

-Solve for Molarity 1

       Molarity 1 = Molarity 2 x Volume 2 / Volume 1

-Substitution

        Molarity 1 = (0.5)(150) / 60

-Simplification

        Molarity = 75 / 60

-Result

        Molarity = 1.25 M

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4 years ago
A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma
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Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:

  • 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
  • n = 0.307 mol

So now we know that

  • MolH₂ + MolN₂ = 0.307 mol

and

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
  • (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
  • 0.614 - 2MolN₂ + 28molN₂ = 3.49
  • 0.614 + 26MolN₂ = 3.49
  • MolN₂ = 0.111 mol

Now we calculate MolH₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ + 0.111 = 0.307
  • MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:

  • N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
  • H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

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Explanation:

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HF (aq) →  H⁺ (aq)  +  F⁻ (aq)

7 0
3 years ago
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