What volume of 0.25 mol/L solution of lead(II)nitrate would be required to form 500mL of a 0.15 mol/L solution of lead(II)nitrat e. Show calculations
1 answer:
To solve this we use the equation,
<span>M1V1 = M2V2</span>
<span>where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.</span>
0.25<span> M x V1 = 0.15 M x .500 L</span>
<span>
V1 = 0.3 L or 300 mL </span>
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