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=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 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=12.82
<u><em>12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution</em></u>
pH=4.625
The classification of this sample of saliva : acid
<h3>Further explanation</h3>
The water equilibrium constant (Kw) is the product of concentration
the ions:
Kw = [H₃O⁺] [OH⁻]
Kw value at 25° C = 10⁻¹⁴
It is known [OH-] = 4.22 x 10⁻¹⁰ M
then the concentration of H₃O⁺:
![\tt 10^{-14}=4.22\times 10^{-10}\times [H_3O^+]\\\\(H_3O^+]=\dfrac{10^{-14}}{4.22\times 10^{-10}}=2.37\times 10^{-5}](https://tex.z-dn.net/?f=%5Ctt%2010%5E%7B-14%7D%3D4.22%5Ctimes%2010%5E%7B-10%7D%5Ctimes%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%28H_3O%5E%2B%5D%3D%5Cdfrac%7B10%5E%7B-14%7D%7D%7B4.22%5Ctimes%2010%5E%7B-10%7D%7D%3D2.37%5Ctimes%2010%5E%7B-5%7D)
pH=-log[H₃O⁺]
Saliva⇒acid(pH<7)
Density= Mass/Volume I am positive I just had an assignment on this