Verified Answer :

\(\boxed{{\text{5}}{\text{.292 g}}}\) 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  \(\left(M\right)\)

2. Molality  \(\left(m\right)\)

3. Mole fraction  \(\left(X\right)\)

4. Parts per million  \(\left({{\text{ppm}}}\right)\)

5. Mass percent  \(\left({\left({{\text{w/w}}}\right)\%}\right)\)

6. Volume percent  \(\left({\left({{\text{v/v}}}\right)\%}\right)\)

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:

\({\text{Molarity of KOH solution}}=\frac{{{\text{amount}}\;\left({{\text{mol}}}\right)\;{\text{of}}\;{\text{KOH}}}}{{\;{\text{volume }}\left({\text{L}}\right)\;{\text{of}}\;{\text{solution}}}}\)                         ......(1)

Rearrange equation (1) to calculate the amount of KOH.

\({\text{Amount}}\;\left({{\text{mol}}}\right)\;{\text{of}}\;{\text{KOH}}=\left( {{\text{Molarity of KOH solution}}}\right)\left({\;{\text{Volume }}\left({\text{L}}\right)\;{\text{of KOH }}{\text{solution}}}\right)\)    ......(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)

\(\begin{aligned}{\text{Amount}}\;\left({{\text{mol}}}\right)\;{\text{of}}\;{\text{KOH}}&=\left({{\text{0}}{\text{.189 mol/L}}}\right)\left({500\;{\text{mL}}}\right)\left({\frac{{1\;{\text{L}}}}{{{{10}^{ - 3}}{\text{mL}}}}}\right)\\&=0.0945\;{\text{mol}}\\\end{aligned}\)

The formula to calculate the mass of KOH is,

\({\text{Mass of KOH}}=\left({{\text{Amount of KOH}}}\right)\left({{\text{Molar mass of KOH}}}\right)\)     ......(3)

The amount of KOH is 0.0945 mol.

The molar mass of KOH is 56 g/mol.

Substitute these values in equation (3)

\(\begin{aligned}{\text{Mass of KOH}}&=\left({{\text{0}}{\text{.0945 mol}}}\right)\left({\frac{{56\;{\text{g}}}}{{1\;{\text{mol}}}}}\right)\\&=5.292\;{\text{g}}\\\end{aligned}\)

So the mass of KOH is 5.292 g.

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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:

\(Molarity=\frac{amount of moles of solute}{volume}\)

Molarity is expressed in units\(\frac{moles}{L}\) .


In this case, you know:

  • molarity= 0.189 M
  • number of moles of solute= ?
  • volume= 500 mL= 0.500 L (being 1000 mL=1 L=

Replacing in the definition of molarity:

\(0.189 M=\frac{amount of moles of solute}{0.5 L}\)

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:

\(mass=0.0945 molesx\frac{56 grams}{1 mole}\)

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