Determine the concentration of a solution prepared by diluting 25.0 mL of a stock 0.188 M Sr(NO₃)₂ solution to 150.0 mL. A) 1.13 M B) 0.0887 M C) 0.0313 M D) 0.0199 M E) 0.0501 M

Question

Quizplus · Accepted Answer

Use the dilution formula:

M₁V₁ = M₂V₂

Where: 
M₁ = initial concentration (stock solution) 
V₁ = initial volume (25.0 mL) 
M₂ = final concentration (unknown) 
V₂ = final volume (150.0 mL)

Substituting:

(0.188 M)(25.0 mL) = M₂(150.0 mL)

M₂ = (0.188 M x 25.0 mL) / 150.0 mL

M₂ = 0.0313 M

Therefore, the final concentration of the diluted solution is 0.0313 M. The correct answer is C.

Determine the Concentration of a Solution Prepared by Diluting 25