Solveeit Logo
Login

Question

Question: How many grams of potassium carbonate \( {K_2}C{O_3} \) are needed to make \( 200{\text{ }}mL \) of ...

How many grams of potassium carbonate K2CO3{K_2}C{O_3} are needed to make 200 mL200{\text{ }}mL of a 2.5 M2.5{\text{ }}M solution?

Explanation

Solution

The mole concept is very significant and useful in chemistry. It is actually the base of stoichiometry and it provides the best option to express the amounts of reactants as well as products that are consumed and formed during a chemical reaction.

Complete answer:
A solution's molarity tells you about the number of moles of solute that is present in 1 L of solution. You can say:
Molarity(M)=moles of solute1 litre of solution Molarity(M) = \dfrac{{moles{\text{ }}of{\text{ solute}}}}{{1{\text{ }}litre{\text{ of }}solution{\text{ }}}}
We know that in order to calculate a solution's molarity, you should know the number of moles of solute that is present in 1 L i.e. 1000 mL of solution.
We can use the molar mass of a compound to determine the number of moles present in the sample so number of moles of a compound can be calculated by applying the following formula:
Number of moles=Mass(g)Molar mass(gmol−1)Number{\text{ }}of{\text{ }}moles = \dfrac{{Mass(g)}}{{Molar{\text{ }}mass(gmo{l^{ - 1}})}}
In the given question, we already know the molarity as well as volume of the solution which are as follows:
\begin{array}{*{20}{l}} {Molarity = 2.5} \\\ {Volume = 200mL = 0.2L} \end{array}
Substituting these given values in the formula of molarity, we will calculate the number of moles as depicted below:
2.5=moles of solute0.2  number of moles=0.5 moles of K2CO3  2.5 = \dfrac{{moles{\text{ }}of{\text{ }}solute}}{{0.2{\text{ }}}} \\\ number{\text{ }}of{\text{ }}moles = 0.5{\text{ }}moles{\text{ }}of{\text{ }}{K_2}C{O_3} \\\
Molar mass of K2CO3{K_2}C{O_3} can be calculated by adding the atomic mass of each of the atom present in the molecule as shown below:
Molecular mass of K2CO3=(2×K)+(1×C)+(3×O)=(2×39)+(1×12)+(3×16)=138gmol−1Molecular{\text{ }}mass{\text{ }}of{\text{ }}{K_2}C{O_3} = (2 \times K) + (1 \times C) + (3 \times O) = (2 \times 39) + (1 \times 12) + (3 \times 16) = 138gmo{l^{ - 1}}
Now substituting the values in the number of moles formula, we get the mass of K2CO3{K_2}C{O_3} as shown below:
0.5=Mass(g)138 Mass(g)=69g \begin{gathered} 0.5 = \dfrac{{Mass(g)}}{{138}} \\\ Mass(g) = 69g \\\ \end{gathered}
Hence, 69 grams of potassium carbonate K2CO3{K_2}C{O_3} are needed to make 200 mL200{\text{ }}mL of a 2.5 M2.5{\text{ }}M solution.

Note:
Don’t get confused between molarity and molality. Molarity is the number of moles of solute per volume of solution (litres) while Molality is the number of moles of solute per weight of solvent (kilogram). These two have a relationship as below:
Molality = (density of the solution − molarity) × molecular weight of solute molarity.