Answer:
48.6 mL
Explanation:
Considering
[tex]Moles\ of\ the\ solution=Molarity_{stock\ solution}\times Volume_{stock\ solution}[/tex]
Given that:
Moles of the solution to be made = 0.00510 mol
[tex]Volume_{stock\ solution}=?[/tex]
[tex]Molarity_{stock\ solution}=0.105\ M[/tex]
So,
[tex]0.00510\ mol=0.105\ M\times Volume_{stock\ solution}[/tex]
[tex]Volume_{stock\ solution}=\frac{0.00510}{0.105}\ L=0.0486\ L[/tex]
The volume of 0.105M stock solution added = 0.0486 L
1 L = 1000 mL
So, volume needed = 48.6 mL
The volume of 0.105 M AgNO3 needed for an experiment that requires 0.00510 mol of AgNO3 is 48.6mL.
The volume of a solution can be calculated using the following expression:
Molarity = no of moles ÷ volume
Volume = no of moles/molarity
According to this question, 0.105 M AgNO3 are needed for an experiment that requires 0.00510 mol of AgNO3.
Volume of AgNO3 = 0.00510 ÷ 0.105
Volume = 0.0486L = 48.6mL
Therefore, the volume of 0.105 M AgNO3 needed for an experiment that requires 0.00510 mol of AgNO3 is 48.6mL.
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