Respuesta :
we can use the following formula when preparing diluted solutions from stock solutions
c1v1 = c2v2
c1 is concentration and v1 is volume of stock solution taken to prepare the diluted solution
c2 is concentration and v2 the volume of the diluted solution
13 M x 51 mL = 0.126 M x V
V = 5262 mL
the final volume of the diluted solution is 5262 mL.
then 51 mL should be added to 5211 mL to get a final volume of 5262 mL diluted solution
c1v1 = c2v2
c1 is concentration and v1 is volume of stock solution taken to prepare the diluted solution
c2 is concentration and v2 the volume of the diluted solution
13 M x 51 mL = 0.126 M x V
V = 5262 mL
the final volume of the diluted solution is 5262 mL.
then 51 mL should be added to 5211 mL to get a final volume of 5262 mL diluted solution
51 mL [tex]{\text{HN}}{{\text{O}}_{\text{3}}}[/tex] solution is diluted to [tex]\boxed{{\text{5210}}{\text{.9 mL}}}[/tex]to get the 0.126 M of solution.
Further explanation:
Molarity:
The molarity of the solution is defined as the concentration of the solution. It is equal to the number of moles of the solute dissolved in one liter of the solution.
The formula of molarity (M), volume (V), and number of moles (n) is as follows:
[tex]{\text{M}} = \frac{{\text{n}}}{{\text{V}}}[/tex] …… (1)
Here V is a volume of solution in liters and n is a number of moles of solute.
If n is constant and is equal to k, the equation (1) will become:
[tex]{\text{MV}} = {\text{k}}[/tex] …... (2)
If we have initial and final values of concentration and volume then the equation (2) can be modified as follows:
[tex]{{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}={{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}[/tex] ……. (3)
Here,
[tex]{{\text{M}}_1}[/tex]is the concentration before dilution.
[tex]{{\text{M}}_2}[/tex]is the concentration after dilution.
[tex]{{\text{V}}_1}[/tex]is the volume before dilution.
[tex]{{\text{V}}_2}[/tex]is the volume after dilution.
On rearranging equation (3) for [tex]{{\text{V}}_{\text{2}}}[/tex] we get,
[tex]{{\text{V}}_{\text{2}}}=\frac{{{{\text{M}}_1}{{\text{V}}_1}}}{{{{\text{M}}_{\text{2}}}}}[/tex] …… (4)
[tex]{{\text{M}}_1}[/tex]is 13 M.
[tex]{{\text{M}}_2}[/tex]is 0.126 M.
[tex]{{\text{V}}_1}[/tex]is 51 mL.
Substitute the values of [tex]{{\text{M}}_1}[/tex], [tex]{{\text{M}}_2}[/tex]and [tex]{{\text{V}}_1}[/tex]in equation(4).
[tex]\begin{aligned}{{\text{V}}_{\text{2}}}&=\frac{{\left({{\text{13 M}}}\right)\times\left({{\text{51 mL}}}\right)}}{{{\text{(0}}{\text{.126}}\;{\text{M)}}}}\\&={\text{5261}}{\text{.90 mL}}\\\end{aligned}[/tex]
The volume of solution after dilution is 5261.90 mL.
The volume added to initial volume of solution is calculated as follows:
[tex]\begin{aligned}{{\text{V}}_{{\text{added}}}}&={{\text{V}}_{\text{2}}}-{{\text{V}}_{\text{1}}} \\&={\text{5261}}{\text{.90 mL}}-{\text{51}}\;{\text{mL}}\\&={\text{5210}}{\text{.9}}\;{\text{mL}}\\\end{aligned}[/tex]
Hence, 51 mL is diluted to 5210.9 mL to get the 0.126 M of solution.
Learn more:
1. Write the equation to show how cation acts as an acid https://brainly.com/question/2396649.
2. Calllculation of heat of vaporization: https://brainly.com/question/10643036
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole concept
Keywords: Number of moles, concentration, volume, volume of solution after dilution, solute, 1liter, molarity, 51 mL, 5210.9 mL, 13m, 0.126m.
