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SSSSS [86.1K]
3 years ago
7

You wish to make a 0.299 M hydroiodic acid solution from a stock solution of 6.00 M hydroiodic acid. How much concentrated acid

must you add to obtain a total volume of 50.0 mL of the dilute solution?
Chemistry
1 answer:
Art [367]3 years ago
5 0

Answer:

V_1=2.49mL

Explanation:

Hello,

In this case, considering that the moles of hydrioiodic acid remain unchanged during the dilution process:

n_{HI}=n_{HI}

One could apply the following equality in terms of molarity:

V_1M_1=V_2M_2

Whereas the subscript 1 accounts for the solution before the dilution and 2 after the dilution, therefore, the required volume of 6.00 M acid is:

V_1=\frac{V_2M_2}{M_1} =\frac{50.0mL*0.299M}{6.00M}=2.49mL

Best regards.

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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
-BARSIC- [3]

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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