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Natasha_Volkova [10]
3 years ago
9

Calculate the volume of a 2.30 mol/L solution of Na2CO3 that contains 0.325 mol of solute.

Chemistry
1 answer:
Blababa [14]3 years ago
5 0

Answer:

141 mL is the volume of a 2.30 mol/L solution of Na₂CO₃ that contains 0.325 mol of solute.

Explanation:

Molarity is a measure of concentration of a solution that indicates the amount of moles of solute that appear dissolved in each liter of the mixture.

So 2.30 mol / L indicates that 2.30 moles dissolve in 1 L of solution. You want to know in what volume is contained 0.325 mol of solute. For this you apply the following rule of three: if 2.30 moles of solute are present in 1 L of solution, 0.325 moles in how much volume is it?

volume=\frac{0.325 moles*1L}{2.30 moles}

volume= 0.141 L

Being 1L=1000 mL, then 0.141 L=141 mL

<u><em>141 mL is the volume of a 2.30 mol/L solution of Na₂CO₃ that contains 0.325 mol of solute.</em></u>

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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
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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
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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:

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  • 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
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Now we calculate MolH₂:

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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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For example: Formation of water from hydrogen gas and oxygen gas. Reaction follows:

2H_2(g)+O_2(g)\rightarrow 2H_2O(l)

The Bonds of hydrogen and oxygen molecule are broken and new bonds between hydrogen and oxygen atoms are formed to give water molecule.

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