
A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCl is needed?
Answer
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Hint: In the given question, a solution of lower concentration 6 M is formed by the method of dilution from the solution of higher concentration 12 M. The relation is given by the equation: ${M_1}{V_1} = {M_2}{V_2}$.
Complete step by step answer:
The volume of HCl solution 2 is 0.500 L
The molarity of HCl solution 2 is 6 M
The molarity of HCl solution 1 is 12 M.
In this question 6 M HCl solution is prepared from 12 M HCl solution by the method of dilution.
The dilution is defined as the method where a solution of lower concentration is prepared from the solution of higher concentration by addition of extra solvent.
The formula of dilution is shown below.
${M_1}{V_1} = {M_2}{V_2}$
Where,
${M_1}$ is the molarity of HCl solution 1.
${V_1}$ is the volume of HCl solution 1.
${M_2}$ is the molarity of HCl solution 2.
${V_2}$ is the volume of HCl solution 2.
To calculate the volume required of a 12 M HCl solution, substitute the values in the above equation.
$ \Rightarrow 12M \times {V_1} = 6M \times 0.500L$
$ \Rightarrow {V_1} = \dfrac{{6M \times 0.500L}}{{12M}}$
$ \Rightarrow {V_1} = 0.25$
Therefore, 0.25 L of HCl solution of 12 M is required to prepare 0.500 L of 6 M HCl solution.
Note:
We can also use the relation of molarity which is equal to moles per liter. As volume and molarity of solution 2 is given, moles can be calculated which can be substituted to find the volume of solution 1. Dilution is only used when lower concentration of solution is formed from higher concentration of solution.
Complete step by step answer:
The volume of HCl solution 2 is 0.500 L
The molarity of HCl solution 2 is 6 M
The molarity of HCl solution 1 is 12 M.
In this question 6 M HCl solution is prepared from 12 M HCl solution by the method of dilution.
The dilution is defined as the method where a solution of lower concentration is prepared from the solution of higher concentration by addition of extra solvent.
The formula of dilution is shown below.
${M_1}{V_1} = {M_2}{V_2}$
Where,
${M_1}$ is the molarity of HCl solution 1.
${V_1}$ is the volume of HCl solution 1.
${M_2}$ is the molarity of HCl solution 2.
${V_2}$ is the volume of HCl solution 2.
To calculate the volume required of a 12 M HCl solution, substitute the values in the above equation.
$ \Rightarrow 12M \times {V_1} = 6M \times 0.500L$
$ \Rightarrow {V_1} = \dfrac{{6M \times 0.500L}}{{12M}}$
$ \Rightarrow {V_1} = 0.25$
Therefore, 0.25 L of HCl solution of 12 M is required to prepare 0.500 L of 6 M HCl solution.
Note:
We can also use the relation of molarity which is equal to moles per liter. As volume and molarity of solution 2 is given, moles can be calculated which can be substituted to find the volume of solution 1. Dilution is only used when lower concentration of solution is formed from higher concentration of solution.
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