Question

A chemist has 4.43g of lithium carbonate and must make a 0.0600 mol/L solution. What volume of solution can be made

Answer 1

Our solute is lithium carbonate ( Li2CO3 ) whose molar mass is 73.9 g/mol.

Procedure:

`\begin{gathered} Molarity\text{ (mol/L) = }\frac{Mass\text{ Li2CO3}}{\text{Molar mass Li2CO3 x Volume (L)}} \\ \text{Volume (L) = }\frac{Mass}{\text{Molar mass x Molarity}}=\frac{4.43\text{ g}}{73.9\text{ }\frac{\text{g}}{mol}x0.0600\text{ }(mol)/(L)}=0.999\text{ L} \\ \end{gathered}`

Answer: Volume = 0.999 L = 1.000 L