Respuesta :
Answer:
A) 0,679 mol/L
B) 10 mol/L
C) 0.07000 mol/L
D) 3.97 mol/L
E) 0.070 mol/L
F) 6.6 mol/L
Explanation:
A) Molarity is the ration between the number of moles ins solution over the solution volume in liters.
Therefore the molarity of such solution is:
[tex]M=\frac{n}{L}=\frac{0.444}{0.654} =0.679[/tex]
B) Upon consulting a periodic table one can find the atomic masses of each element present at a given molecule.
H=1.00 g/mol P=30.97 O=16.00
By adding the masses of the atoms in a molecule, we get its molecular weight.
H3P04 molecular mass is 3x1.00+30.97+4x16.00=97.97g/mol
The molar mass is used to calculate the number of moles of a molecule in a given sample.
[tex]n=\frac{sample}{molar mass} = \frac{98.0}{97.97} =1.00[/tex]
Therefore the molarity of such solution is:
[tex]M=\frac{n}{L}=\frac{1}{0.1} =10[/tex]
C) Upon consulting a periodic table one can find the atomic masses of each element present at a given molecule.
H=1.00 g/mol Ca=40.08 O=16.00
By adding the masses of the atoms in a molecule, we get its molecular weight.
Ca(OH)2 molecular mass is 2x1.00+40.08+2x16.00=74.08g/mol
The molar mass is used to calculate the number of moles of a molecule in a given sample.
[tex]n=\frac{sample}{molar mass} = \frac{0.2074}{74.08} =0.002800[/tex]
Therefore the molarity of such solution is:
[tex]M=\frac{n}{L}=\frac{0.002800}{0.0040} =0.07000[/tex]
D) Upon consulting a periodic table one can find the atomic masses of each element present at a given molecule. (assuming you meant Na2SO4)
Na=22.99 g/mol S=32.07 O=16.00
By adding the masses of the atoms in a molecule, we get its molecular weight.
Na2SO4 molecular mass is 2x22.99+32.07+4x16.00=142.05 g/mol
The molar mass is used to calculate the number of moles of a molecule in a given sample.
[tex]n=\frac{sample}{molar mass} = \frac{10500}{142.05} =73.9[/tex]
Therefore the molarity of such solution is:
[tex]M=\frac{n}{L}=\frac{73.9}{18.60} =3.97[/tex]
E) Molarity is the ration between the number of moles ins solution over the solution volume in liters.
Therefore the molarity of such solution is:
[tex]M=\frac{n}{L}=\frac{0.0070}{0.1000} =0.070[/tex]
F) Upon consulting a periodic table one can find the atomic masses of each element present at a given molecule.
H=1.00 g/mol Cl=35.45
By adding the masses of the atoms in a molecule, we get its molecular weight.
HCl molecular mass is 1.00+35.45=36.45g/mol
The molar mass is used to calculate the number of moles of a molecule in a given sample.
[tex]n=\frac{sample}{molar mass} = \frac{18}{36.45} =0.49[/tex]
Therefore the molarity of such solution is:
[tex]M=\frac{n}{L}=\frac{0.49}{0.075} =6.6[/tex]
Considering the definition of molarity and mass molar, you obtain:
(a) The molarity of the solution is 0.68[tex]\frac{moles}{L}[/tex].
(b) The molarity of the solution is 0.01[tex]\frac{moles}{L}[/tex].
(c) The molarity of he solution is 0.07[tex]\frac{moles}{L}[/tex].
(d) The molarity of he solution is 1.75[tex]\frac{moles}{L}[/tex].
(e) The molarity of he solution is 0.07[tex]\frac{moles}{L}[/tex].
(f) The molarity of he solution is 6.53[tex]\frac{moles}{L}[/tex].
Molar concentration or molarity is a measure of the concentration of a solute in a solution and is the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the number of moles of the solute by the volume of the solution:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{L}[/tex].
(a) In this case you have 0.444 moles of COCl₂ in 0.654 L of solution. Then, the molarity can be calculated as:
[tex]Molarity=\frac{0.444 moles}{0.654 L}[/tex]
Solving:
Molarity= 0.68[tex]\frac{moles}{L}[/tex]
The molarity of the solution is 0.68[tex]\frac{moles}{L}[/tex].
(b) In this case you have 98.0 gram of phosphoric acid, H₃PO₄, in 100 L of solution. Then, being the molar mass of phosphoric acid 98 g/mole, the number of moles that 98 grams of the compound contain is 1.
Then, the molarity can be calculated as:
[tex]Molarity=\frac{1 moles}{100 L}[/tex]
Solving:
Molarity= 0.01[tex]\frac{moles}{L}[/tex]
The molarity of the solution is 0.01[tex]\frac{moles}{L}[/tex].
(c) In this case you have 0.2074 g of calcium hydroxide, Ca(OH)₂, in 40.00 mL of solution. Then, being the molar mass of calcium hydroxide 74 g/mole, the number of moles that 0.2074 grams of the compound can be calculated as:
[tex]number of moles=0.2074 gramsx\frac{1 mole}{74 grams}[/tex]
Solving:
number of moles= 2.80×10⁻³ moles
On the other side, the volume is 40 mL= 0.04 L (being 1000 mL= 1 L)
Then, the molarity can be calculated as:
[tex]Molarity=\frac{2.80x10^{-3} moles}{0.04 L}[/tex]
Solving:
Molarity= 0.07[tex]\frac{moles}{L}[/tex]
The molarity of he solution is 0.07[tex]\frac{moles}{L}[/tex].
(d) In this case you have 10.5 kg =10500 g, (being 1 kg=1000 g) of Na₂SO₄.10H₂O in 18.60 L of solution. Then, being the molar mass of the compound 322 g/mole, the number of moles that 10500 grams of the compound can be calculated as:
[tex]number of moles=10500 gramsx\frac{1 mole}{322 grams}[/tex]
Solving:
number of moles= 32.61 moles
Then, the molarity can be calculated as:
[tex]Molarity=\frac{32.61 moles}{18.60 L}[/tex]
Solving:
Molarity= 1.75[tex]\frac{moles}{L}[/tex]
The molarity of he solution is 1.75[tex]\frac{moles}{L}[/tex].
(e) In this case you have 7.0x10⁻³ moles of l₂ in 100.0 ml of solution. Then, the volume is 100 mL= 0.1 L (being 1000 mL= 1 L)
Then, the molarity can be calculated as:
[tex]Molarity=\frac{7.0x10^{-3} moles}{0.1 L}[/tex]
Solving:
Molarity= 0.07[tex]\frac{moles}{L}[/tex]
The molarity of he solution is 0.07[tex]\frac{moles}{L}[/tex].
(f) In this case you have 1.8x10⁴ mg= 18 g (being 1 mg=0.001 g) of HCl in 0.075 L of solution. Then, being the molar mass of HCl 36.45 g/mole, the number of moles that 18 grams of the compound can be calculated as:
[tex]number of moles=18 gramsx\frac{1 mole}{36.45 grams}[/tex]
Solving:
number of moles= 0.49 moles
Then, the molarity can be calculated as:
[tex]Molarity=\frac{0.49 moles}{0.075 L}[/tex]
Solving:
Molarity= 6.53[tex]\frac{moles}{L}[/tex]
The molarity of he solution is 6.53[tex]\frac{moles}{L}[/tex].
Learn more about molarity with this example: brainly.com/question/15406534?referrer=searchResults
