Respuesta :
Answer:
0.40 L
Explanation:
Calculation of the moles of [tex]Ba(OH)_2[/tex] as:-
Mass = 51.24 g
Molar mass of [tex]Ba(OH)_2[/tex] = 171.34 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{51.24\ g}{171.34\ g/mol}[/tex]
[tex]Moles= 0.2991\ mol[/tex]
Volume = 1.20 L
The expression for the molarity is:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{0.2991\ mol}{1.20\ L}=0.24925\ M[/tex]
Thus,
Considering
[tex]Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}[/tex]
Given that:
[tex]Molarity_{working\ solution}=0.100\ M[/tex]
[tex]Volume_{working\ solution}=1\ L[/tex]
[tex]Volume_{stock\ solution}=?[/tex]
[tex]Molarity_{stock\ solution}=0.24925\ M[/tex]
So,
[tex]0.100\ M\times 1\ L=0.24925\ M\times Volume_{stock\ solution}[/tex]
[tex]Volume_{stock\ solution}=\frac{0.100\times 1}{0.24925}\ L=0.40\ L[/tex]
The volume of 0.24925M stock solution added = 0.40 L
