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9. What volume of 0.416M Mg(NO3)2 should be added to 255 mL of 0.102 M KNO3 to produce a solution with a concentration of 0.278 M NO3– ions? Assume volumes are additive.

Question

9. What volume of 0.416M Mg(NO3)2 should be added to 255 mL of 0.102 M KNO3 to produce a solution with a concentration of 0.278 M NO3 ions? Assume volumes are additive. 

 

 

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Step 1

When two solutions are mixed with each other, then dilution equation can be represented as follows:

M,V+M,VMV
Here
M Molarity of the solution
V =Volume in liters
Step-2
Given:
Concentration of Mg (NO, =0.416 M
Concentration of KNO3 0.102 M
Volume 255 mL
Concentrati on of NO 0.278 M
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M,V+M,VMV Here M Molarity of the solution V =Volume in liters Step-2 Given: Concentration of Mg (NO, =0.416 M Concentration of KNO3 0.102 M Volume 255 mL Concentrati on of NO 0.278 M

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Step 2

The values are given as follows:

Concentration of Mg (NO), =0.416 M
Concentration of KNO, 0.102 M
Volume 255 mL
Concentrati on of NO
0.278 M
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Concentration of Mg (NO), =0.416 M Concentration of KNO, 0.102 M Volume 255 mL Concentrati on of NO 0.278 M

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Step 3

Suppose the unknown volume is X and as 2 moles of NO3 is present in Mg (NO3)2,...

2x0.416 M 0.832 M
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2x0.416 M 0.832 M

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