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vodomira [7]
3 years ago
15

What would be the final volume of the new solution if the 0.2 m solution on the left were diluted to 0.04 m? ml quizley?

Chemistry
1 answer:
marin [14]3 years ago
3 0
Missing question: volume of <span>solution on the left is 10 mL.
V</span>₁(solution) = 10 Ml.
c₁(solution) = 0.2 M.<span>
V</span>₂(solution) = ?.<span>
c</span>₂(solution) = 0.04 M.<span>
c</span>₁ - original concentration of the solution, before it gets diluted.<span>
c</span>₂ - final concentration of the solution, after dilution.<span>
V</span>₁ - <span>volume to be diluted.
V</span>₂ - <span>final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
</span>10 mL · 0.2 M = 0.04 M · V₂.
V₂(solution) = 10 mL · 0.2 M  ÷ 0.04 M.
V₂(solution) = 50 mL.<span>

</span>
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<h3>Answer:</h3>

300 kPa

<h3>Explanation:</h3>

Concept tested: Gas Laws (Combined gas law)

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Initial temperature,T₁ = 27°C

But, K = °C + 273, therefore, T₁ = 300 K

Initial volume, V₁ = 900 L

New temperature, T₂ = -73°C

                                   = -73°C + 273

                                   = 200 K

New volume, V₂ = 200 L             

We are required to calculate the new pressure, P₂

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  • According to the combined gas law;

P₁V₁/T₁ = P₂V₂/T₂

  • To calculate the new Pressure, P₂, we rearranging the formula;

P₂ =(P₁V₁T₂) ÷ (V₂T₁)

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P₂ = (100 kPa × 900 L × 200 K) ÷ (200 L × 300 K)

   = 300 kPa

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