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Fed [463]
3 years ago
14

In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How

many moles of CaF2 can dissolve in 100 g of water? If the density of a saturated solution of CaF2 is 1.00 g/mL, how many moles of CaF2 will dissolve in exactly 1.00 L of solution?
Chemistry
1 answer:
____ [38]3 years ago
5 0

Answer:

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution

Explanation:

First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:

  • Ca: 40 g/mole
  • F: 19 g/mole

So the molar mass of CaF₂ is:

CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?

moles=\frac{0.0016 grams*1 mole}{78 grams}

moles=2.05*10⁻⁵

<u><em>2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.</em></u>

Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?

mass of CaF_{2}=\frac{1000 mL*1g}{1mL}

mass of CaF₂= 1000 g

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?

moles=\frac{1000 grams*1 mole}{78 grams}

moles=12.82

<u><em>12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution</em></u>

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The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

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<u>Explanation:</u>

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