Final answer:
The concentration of a solution prepared by diluting 25.0 mL of 0.188M Sr(NO3)2 to 150.0 mL is determined using the dilution equation C1V1 = C2V2. After calculation, the final molarity of the diluted solution is found to be 0.0313M.
Step-by-step explanation:
To determine the concentration of a solution prepared by diluting 25.0 mL of a stock 0.188M Sr(NO3)2 solution to 150.0 mL, we use the dilution equation, which states that the concentration of the stock solution multiplied by its volume (C1V1) will equal the concentration of the diluted solution multiplied by its volume (C2V2). Thus, we have:
C1V1 = C2V2
Where:
- C1 is the concentration of the stock solution (0.188M)
- V1 is the volume of the stock solution before dilution (25.0 mL)
- C2 is the concentration of the solution after dilution (this is what we're solving for)
- V2 is the final volume of the diluted solution (150.0 mL)
To find C2, we rearrange the equation:
C2 = (C1V1) / V2
And enter the values:
C2 = (0.188M * 25.0 mL) / 150.0 mL
C2 = (4.7 mM) / 150.0 mL
C2 = 0.03133M or 0.0313M when rounding to three significant figures.
The concentration of the diluted Sr(NO3)2 solution is therefore 0.0313M after dilution.