Complete Question
A chemist prepares a solution of barium chlorate BaClO32 by measuring out 42.g of barium chlorate into a 500.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /molL of the chemist's barium chlorate solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration is [tex]C = 0.28 \ mol/L[/tex]
Explanation:
  From the question we told that
   The mass of [tex]Ba(ClO_{3})_2[/tex] is  [tex]m_b = 42 \ g[/tex]
   The volume of the solution  [tex]V_s = 500 mL = 500*10^{-3} L[/tex]
Now the number f moles of  [tex]Ba(ClO_{3})_2[/tex] in the solution is mathematically represented as
    [tex]n = \frac{m_b}{Z_b}[/tex]
Where  [tex]Z_b[/tex] is the molar mass of [tex]Ba(ClO_{3})_2[/tex] which a constant with a value
       [tex]Z_b = 304.23 \ g/mol[/tex]
Thus
    [tex]n = \frac{42}{304.23}[/tex]
   [tex]n = 0.14 \ mol[/tex]
The concentration of [tex]Ba(ClO_{3})_2[/tex] Â in the solution is mathematically evaluated as
    [tex]C = \frac{n}{V_2}[/tex]
substituting  values Â
   [tex]C = \frac{0.14}{500*10^{-3}}[/tex]
   [tex]C = 0.28 \ mol/L[/tex]
