Answer: The mass concentration for a solution containing 45 g of calcium carbonate for 100 [tex]cm^{3}[/tex] is 0.045 M.
Explanation:
Given: Mass of calcium carbonate = 45 g
Volume = [tex]100 cm^{3}[/tex]
Convert [tex]cm^{3}[/tex] into L as follows.
[tex]1 cm^{3} = 0.001 L\\100 cm^{3} = 100 cm^{3} \times \frac{0.001 L}{1 cm^{3}}\\= 0.1 L[/tex]
Moles of calcium carbonate (molar mass = 100 g/mol) is as follows.
[tex]No. of moles = \frac{45 g}{100 g/mol}\\= 0.45 mol[/tex]
Molarity is the number of moles of solute present in a liter of solution.
Hence, molarity (or mass concentration) of given solution is calculated as follows.
[tex]Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.45 mol}{0.1 L}\\= 0.045 M[/tex]
Thus, we can conclude that the mass concentration for a solution containing 45 g of calcium carbonate for 100 [tex]cm^{3}[/tex] is 0.045 M.
