Respuesta :
Answer : The volume of 4.9 M [tex]H_2O_2[/tex] stock solution used to prepare the solution is, 12.24 ml
Solution : Given,
Molarity of aqueous [tex]H_2O_2[/tex] solution = 1.20 M = 1.20 mole/L
Volume of aqueous [tex]H_2O_2[/tex] solution = 50.0 ml = 0.05 L
(1 L = 1000 ml)
Molarity of [tex]H_2O_2[/tex] stock solution = 4.9 M = 4.9 mole/L
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = Molarity of aqueous [tex]H_2O_2[/tex] solution
[tex]M_2[/tex] = Molarity of [tex]H_2O_2[/tex] stock solution
[tex]V_1[/tex] = Volume of aqueous [tex]H_2O_2[/tex] solution
[tex]V_2[/tex] = Volume of [tex]H_2O_2[/tex] stock solution
Now put all the given values in this formula, we get the volume of [tex]H_2O_2[/tex] stock solution.
[tex](1.20mole/L)\times (0.05L)=(4.9mole/L)\times V_2[/tex]
By rearranging the term, we get
[tex]V_2=0.01224L=12.24ml[/tex]
Therefore, the volume of 4.9 M [tex]H_2O_2[/tex] stock solution used to prepare the solution is, 12.24 ml
Answer:
For preparing 50 ml of 1.20 M [tex]\rm H_2O_2[/tex] solution 12.44 ml of 4.9 M solution is required.
Explanation:
For the preparation of working or diluted solution from the stock, the following equation rule is followed:
[tex]\rm M_1V_1\;=\;M_2V_2[/tex]
where [tex]\RM M_1\;and\;V_1[/tex] are molarity and volume of 1st solution,
[tex]\RM M_2\;and\;V_2[/tex] are molarity and volume of the second solution.
Given, Molarity of solution to be prepared [tex]\rm (M_1)[/tex] = 1.20 M
The volume of the solution to be prepared [tex]\rm (V_1)[/tex] = 50.0 ml
Molarity of the given [tex]\rm H_2O_2[/tex] solution [tex]\rm (M_2)[/tex] = 4.9 M
From the equation,
[tex]\rm 1.20\;\times\;50.0\;=\;4.9\;\times\;V_2[/tex]
[tex]\rm V_2[/tex] = 12.24 ml
For the preparation of 50 ml of 1.20 M [tex]\rm H_2O_2[/tex] solution, 12.24 ml of 4.9 M [tex]\rm H_2O_2[/tex] solution has to be dissolved and make up to 50 ml.
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