what volume of 0.25mol/L solution lead (ii) nitrate would be required to form 500ml of a 0.015mol/L solution of lead (ii) nitrate

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Dilution is a method in which a specific amount of solvent is added to a solution so as to decrease its concentration.

It has a general formula of:

M1V1 = M2V2

now we have to assign the values for each

M1 = 0.25 mol/L

M2 = 0.015 mol/L

V1 =...

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Dilution is a method in which a specific amount of solvent is added to a solution so as to decrease its concentration.

It has a general formula of:

M1V1 = M2V2

now we have to assign the values for each

M1 = 0.25 mol/L

M2 = 0.015 mol/L

V1 = ?

V2 = 500 mL = .5 L

The problem asks how much concentrated solution (0.25mol/L) is used to create a 500mL (or .5L) 0.015 mol/L solution. Sounds ok?

We can assume that V1 should be smaller than V2.

then we apply the formula

M1V1= M2V2

(0.25mol/L) (V1) = (0.015mol/L)(.5L)

V1 = 0.03L 

To convert into mL, we just multiply it by 1000.

V1 = 30mL :)

 

 

 

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