The concentration of K extsuperscript{+} ions in 100 mL of 0.0500 M K extsubscript{2}CO extsubscript{3} solution, in g L extsuperscript{-1}, is A - VCE - SSCE Chemistry - Question 12 - 2006 - Paper 1

1. Determine the molarity of K extsubscript{2}CO extsubscript{3}:

   The solution has a molarity of 0.0500 M.

2. Find the number of K extsuperscript{+} ions in K extsubscript{2}CO extsubscript{3}:

   Each formula unit of K extsubscript{2}CO extsubscript{3} yields 2 K extsuperscript{+} ions.

3. Calculate the total moles of K extsuperscript{+} ions:

   Moles of K extsuperscript{+} = 2 imes 0.0500 imes 0.100 = 0.0100 ext{ moles}.

4. Convert moles to grams:

   The molar mass of K = 39.10 g/mol, so:

   Mass of K extsuperscript{+} = 0.0100 imes 39.10 = 0.391 ext{ g}.

5. Find concentration in g L extsuperscript{-1}:

   Since the solution volume is 0.100 L, the concentration of K extsuperscript{+} ions is:

   Concentration = 0.391 ext{ g} ext{ in 1 L} = 0.391 ext{ g L}^{-1}.

Thus, the correct answer is B. 0.391.