[tex]\boxed{{\text{5}}{\text{.292 g}}}[/tex] of KOH is present in 500 mL of 0.189 M KOH solution.
Further Explanation:
The concentration is the proportion of substance in the mixture. The most commonly used concentration terms are as follows:
1. Molarity [tex]\left(M\right)[/tex]
2. Molality [tex]\left(m\right)[/tex]
3. Mole fraction [tex]\left(X\right)[/tex]
4. Parts per million [tex]\left({{\text{ppm}}}\right)[/tex]
5. Mass percent [tex]\left({\left({{\text{w/w}}}\right)\%}\right)[/tex]
6. Volume percent [tex]\left({\left({{\text{v/v}}}\right)\%}\right)[/tex]
Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.
The formula to calculate the molarity of KOH is as follows:
[tex]{\text{Molarity of KOH solution}}=\frac{{{\text{amount}}\;\left({{\text{mol}}}\right)\;{\text{of}}\;{\text{KOH}}}}{{\;{\text{volume }}\left({\text{L}}\right)\;{\text{of}}\;{\text{solution}}}}[/tex] ......(1)
Rearrange equation (1) to calculate the amount of KOH.
[tex]{\text{Amount}}\;\left({{\text{mol}}}\right)\;{\text{of}}\;{\text{KOH}}=\left( {{\text{Molarity of KOH solution}}}\right)\left({\;{\text{Volume }}\left({\text{L}}\right)\;{\text{of KOH }}{\text{solution}}}\right)[/tex] ......(2)
The molarity of the KOH solution is 0.189 M or 0.189 mol/L.
The volume of KOH solution is 500 mL.
Substitute these values in equation (2)
[tex]\begin{aligned}{\text{Amount}}\;\left({{\text{mol}}}\right)\;{\text{of}}\;{\text{KOH}}&=\left({{\text{0}}{\text{.189 mol/L}}}\right)\left({500\;{\text{mL}}}\right)\left({\frac{{1\;{\text{L}}}}{{{{10}^{ - 3}}{\text{mL}}}}}\right)\\&=0.0945\;{\text{mol}}\\\end{aligned}[/tex]
The formula to calculate the mass of KOH is,
[tex]{\text{Mass of KOH}}=\left({{\text{Amount of KOH}}}\right)\left({{\text{Molar mass of KOH}}}\right)[/tex] ......(3)
The amount of KOH is 0.0945 mol.
The molar mass of KOH is 56 g/mol.
Substitute these values in equation (3)
[tex]\begin{aligned}{\text{Mass of KOH}}&=\left({{\text{0}}{\text{.0945 mol}}}\right)\left({\frac{{56\;{\text{g}}}}{{1\;{\text{mol}}}}}\right)\\&=5.292\;{\text{g}}\\\end{aligned}[/tex]
So the mass of KOH is 5.292 g.
Learn more:
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Solutions
Keywords: KOH, molarity, 500 mL, 5.292 g, mol/L, 0.189 M, molar mass, 56 g/mol, concentration, concentration terms, volume.
Considering the definition of molarity and mass molar, the mass of solute in 500.0 mL of 0.189 M KOH is 5.2912 grams.
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{amount of moles of solute}{volume}[/tex]
Molarity is expressed in units[tex]\frac{moles}{L}[/tex] .
In this case, you know:
Replacing in the definition of molarity:
[tex]0.189 M=\frac{amount of moles of solute}{0.5 L}[/tex]
Solving:
number of moles of solute= 0.189 M× 0.5 L
number of moles of solute= 0.0945 moles
Being the molar mass of KOH 56 g/mol, the mass of KOH can be calculated as:
[tex]mass=0.0945 molesx\frac{56 grams}{1 mole}[/tex]
mass= 5.292 grams
Finally, the mass of solute in 500.0 mL of 0.189 M KOH is 5.2912 grams.
Learn more about molarity with this example: brainly.com/question/15406534?referrer=searchResults
