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A chemist prepares a solution of barium chlorate \([tex] Ba \left( ClO_3 \right)_2 \)[/tex] by measuring out \([tex] 99.8 \mu mol \)[/tex] of barium chlorate in a 400 mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in \([tex] mmol/L \)[/tex] of the chemist's barium chlorate solution. Be sure your answer has the correct number of significant digits.

\[ \square \, \frac{mmol}{L} \, \square \, \times 10 \]

Answer :

Sure, let's solve this step by step:

1. Understand the Units:
- The amount of barium chlorate given is 99.8 micromoles (µmol).
- The volume of the solution is 400.0 milliliters (mL).

2. Convert Micromoles to Millimoles:
- To convert from micromoles to millimoles, remember that 1 micromole (µmol) is equal to 0.001 millimoles (mmol).
- So, 99.8 µmol is:
[tex]\[
99.8 \, \mu \text{mol} \times 0.001 = 0.0998 \, \text{mmol}
\][/tex]

3. Convert Milliliters to Liters:
- Since 1 milliliter (mL) is equal to 0.001 liters (L), convert 400.0 mL to liters:
[tex]\[
400.0 \, \text{mL} \times 0.001 = 0.4 \, \text{L}
\][/tex]

4. Calculate the Concentration in Millimoles per Liter (mmol/L):
- The concentration of a solution is given by the formula:
[tex]\[
\text{Concentration} = \frac{\text{amount of solute in mmol}}{\text{volume of solution in L}}
\][/tex]
- Substitute the values we have:
[tex]\[
\text{Concentration} = \frac{0.0998 \, \text{mmol}}{0.4 \, \text{L}} = 0.2495 \, \text{mmol/L}
\][/tex]

Therefore, the concentration of the barium chlorate solution is [tex]\(0.2495 \, \frac{\text{mmol}}{\text{L}}\)[/tex].

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