Answer: the molarity of the solution in volumetric flask "B' is 0.0100 M
Explanation:
Given that;
the Molarity of stock solution M₁ = 1.25M
The molarity os solution in volumetric flask A (M₂) = M₂
Volume of stock solution pipet out (V₁) = 5.00mL
Volume of solution in volumetric flask A V₂ = 25.00mL
using the dilution formula
M₁V₁ = M₂V₂
M₂ = M₁V₁ / V₂
WE SUBSTITUTE
M₂ = ( 1.25 × 5.00 ) / 25.00 mL
M₂ = 0.25 M
Now volume of solution pipet out from volumetric flask A V₂ = 2.00 mL
Molarity of solution in volumetric flask B (M₃) = M₃
Volume of solution in volumetric flask B V₃ = 50.00m L
Using dilution formula again
M₂V₂ = M₃V₃
M₃ = M₂V₂ / V₃
WE SUBSTITUTE
M₃ = ( 0.25 × 2.0) / 50.0
M₃ = 0.0100 M
Therefore the molarity of the solution in volumetric flask "B' is 0.0100 M
The concentration of the final solution is 0.01 M.
This is a problem of serial dilution. We have to first obtain the concentration of the solution in the new flask.
C1V1 = C2 V2
C1 = concentration of stock solution = 1.25 M
V1 = volume of stock solution = 5.00 mL
C2 = concentration of solution in the new flask = ?
V2 = volume of solution in flask B in the new flask = 25.00 mL
C2 = C1V1 /V2
C2 = 1.25 M × 5.00 mL/ 25.00 mL
C2 = 0.25 M
Again we need to find the concentration when this solution is further diluted;
C1 = 0.25 M
V1 = 2.00 mL
C2 = ?
V2 = 50.00 mL
C2 = C1V1/V2
C2 = 0.25 M × 2.00 mL/50.00 mL
C2 = 0.01 M
The concentration of the final solution is 0.01 M.
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