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-BARSIC- [3]
4 years ago
7

Suppose that you have a 60.0% solution of NaOH. How many milliliters of water must be added to 30.0 mL of this solution to prepa

re a 35.0% solution of NaOH?15.5 mL21.4 mL51.4 mL72.8 mLCan’t determine the concentration because you need the density of the solution
Chemistry
1 answer:
defon4 years ago
3 0

Answer:

  • <u>21.4 ml (second choice)</u>

Explanation:

<u>1) Data:</u>

a) C₁ = 60.0% (initial solution)

b) V₁ = 30.0 ml (initial solution)

c) C₂ = 0% (pure water)

d) V₂ = ? (pure water)

e) C₃ = 35.0% (final concentration)

<u>2) Formula:</u>

  • C₁V₁ + C₂V₂ = C₃V₃
  • V₁ + V₂ = V₃ (assuming volume addtivity)

<u>3) Solution:</u>

<u />

a) Substitute values in the first formula:

  • 60.0% × 30 ml + 0 = 35% (30 ml + V₂)

b) Solve the equation (units are supressed just to manipulate the terms)

  • 18 = 10.5 + 0.35V₂

  • 0.35V₂ = 18 - 10.5 = 7.5

  • V₂ = 7.5 / 0.35 = 21.4 ml ← answer    
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Answer:

The type of salt to be added to the water is not known from the question but no worries, I will try to give you the step by step procedure to answer any type of question similar to this.

To answer this question, we should know some facts.

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So we can calculate the molarity of the salt NaCl from the formula;

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m = 5 / 2 * 1.86

m = 5 / 3.72

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number of moles = 1.344 mol/kg * 1 Kg/L * 1 kg water

number of moles = 1.344 moles.

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