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Bess [88]
3 years ago
14

Suppose now that you wanted to determine the density of a small crystal to confirm that it is sulfur. From the literature, you k

now that sulfur has a density of 2.07 . How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of ( = 1.492 g/mL) and ( = 2.890 g/mL)? (Note: 1 mL = 1 .)
Chemistry
1 answer:
ivolga24 [154]3 years ago
7 0

Answer:

Mix 11.73 mL of CHCl₃ and 8.27 mL of CHBr₃

Explanation:

To begin, it seems your question lacks the name of the pure samples that will be mixed to prepare the liquid mixture. The names are not necessary, as the numerical values are there. However, an internet search tells me they are CHCl₃ (d=1.492g/mL) and CHBr₃ (d=2.890 g/mL).

The mixture has a density of 2.07 g/mL, so <u>20 mL of the mixture would weigh</u>:

20 mL * 2.07 g/mL = 41.4 g

Let X be the volume of CHCl₃ and Y the volume of CHBr₃:

X + Y = 20 mL

The mass of CHCl₃ and CHBr₃ combined have to be equal to the mass of the mixture. We can write that equation using the volume of the samples and their density:

X * 1.492 + Y * 2.890 = 41.4 g

So now we have a system of two equations and two unknowns, we use algebra to solve it:

   1. Express Y in terms of X:

X + Y = 20

  • Y = 20 - X

   2. Replace Y in the second equation:

X * 1.492 + Y * 2.890 = 41.4

  • 1.492*X + 2.890*(20-X) = 41.4

   3. Solve for X:

  • 1.492*X + 57.8 - 2.890*X = 41.4
  • 1.398*X = 16.4
  • X = 11.73 mL

   4. Using the now known value of X, solve for Y:

X + Y = 20

  • 11.73 + Y = 20
  • Y = 8.27 mL

So, to prepare the liquid mixture we would mix 8.27 mL of CHBr₃ and 11.73 mL of  CHCl₃.

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Answer:

<u />

  • <u>a) 1.44g</u>

<u />

  • <u>b) 77.3%</u>

<u />

Explanation:

<u>1. Chemical balanced equation (given)</u>

       NaOH(aq)+ KHP(s)\rightarrow NaKP(aq)+H_2O(l)

<u>2. Mole ratio</u>

1molNaOH(aq):1molKHP(s)

This is, 1 mol of NaOH will reacts with 1 mol of KHP.

<u />

<u>3. Find the number of moles in 72.14 mL of the base</u>

    Molarity=\text{number of moles of solute}/\text{volume of solution in liters}

    \text{Volume of solution}=72.14mL=0.07214liters

     \text{Number of moles of NaOH}=0.0978M\times 0.07214liter=0.007055mol

<u>4. Find the number of grams of KHP that reacted</u>

The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol

Convert moles to grams:

  • mass = number moles × molar mass = 0.007055mol × 204.23g/mol
  • mass = 1.4408 g.

You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).

<u>5. Find the percentage of KHP in the sample</u>

The percentage is how much of the substance is in 100 parts of the sample.

The formula is:

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mol of Na2CO3 = 2.36 x 10⁻⁴

<h3>Further explanation</h3>

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