Answer:
[tex]m_{K_2SO_4}=72.84gK_2SO_4[/tex]
Explanation:
Hello,
In this case, by considering that the both the volume of the solution of potassium sulfate and its concentration are given, one solves for the required mass considering the following proportional factors related stoichiometrically:
[tex]m_{K_2SO_4}=550mLsln*\frac{1Lsln}{1000mLsln}*0.76\frac{molK_2SO_4}{Lsln}*\frac{174.259gK_2SO_4}{1molK_2SO_4}\\ m_{K_2SO_4}=72.84gK_2SO_4[/tex]
Best regards.
Taking into account the definition of molarity, you would measure out 72.732 g of K₂SO₄ to prepare 550 ml of a 0.76 M solution.
Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:
[tex]molarity=\frac{number of moles}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liters}[/tex].
In this case, you prepare 550 ml of a 0.76 M solution. This is, you know:
Replacing in the definition of molarity:
[tex]0.76 M=\frac{number of moles}{0.550 L}[/tex]
Solving:
number of moles= 0.76 M × 0.550 L
number of moles= 0.418 moles
Being the molar mass of K₂SO₄ 174 [tex]\frac{g}{mole}[/tex], that is, the amount of mass that the compound contains in one mole, the mass contained in 0.418 moles is calculated as:
0.418 moles×174 [tex]\frac{g}{mole}[/tex]= 72.732 g
Finally, you would measure out 72.732 g of K₂SO₄ to prepare 550 ml of a 0.76 M solution.
Learn more about molarity:
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