A student measures 5 mL of .356 M stock solut'n and pours it into a 250 mL v-metric flask, which is filled to the mark with DI water & mixed?

Method 1 :

The dilution factors in the large and small volumetric flasks are 5/250 and 5/50 respectively.
Concentration of the solution in the small volumetric flask = (0.356 M) × (5/250) × (5/50) = 0.000712 M


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Method 2 :

No. of moles of solute in 5 mL of the stock solution = (0.356 mmol/mL) × (5 mL) = 1.78 mmol
No. of moles of solute in 250 mL solution in the large volumetric flask = 1.78 mmol
No. of moles of solute transfer to the small volumetric flask = (1.78 mmol) × (5/250) = 0.0356 mmol
Concentration of the solution in the small volumetric flask = (0.0356 mmol) / (50 mL) = 0.000712 M


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Method 3 :

Consider the first dilution (using the large volumetric flask) :
C₁ = 0.356 M, V₁ = 5 mL
C₂ = ? M, V₂ = 250 mL

C₁ V₁ = C₂ V₂
(0.356 M) × (5 mL) = C₂ × (250 mL)
Concentration of the solution in the large volumetric flask, C₂ = 0.00712 M

Consider the second dilution (using the small volumetric flask) :
C₃ = 0.00712 M, V₃ = 5 mL
C₄ = ? M, V₄ = 50 mL

C₃ V₃ = C₄ V₄
(0.00712 M) × (5 mL) = C₄ × (250 mL)
Concentration of the solution in the large volumetric flask, C₄ = 0.00712 M