Answer:
9.2375 grams of lithium carbonate is required to prepare 500.0 mL of a 0.250 M Li₂CO₃ solution.
Explanation:
Molarity is a measure of the concentration of a substance and is defined as the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:
[tex]molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case:
Replacing in the definition of molarity:
[tex]0.250 M=\frac{number of moles of solute}{0.5 L}[/tex]
Solving
0.250 M*0.5 L = number of moles of solute
0.125 = number of moles of solute
If the molar mass of lithium carbonate is equal to 73.9 g/mol, then the required mass of the compound can be calculated by:
[tex]73.9 \frac{g}{mol}*0.125 moles= 9.2375 grams[/tex]
9.2375 grams of lithium carbonate is required to prepare 500.0 mL of a 0.250 M Li₂CO₃ solution.
