Question

n empty drying flask has a mass of 28.5 g. A 12.0 mL solution of calcium carbonate is poured into the flask increasing the mass to 38.5 g. When the drying glass is heated so that all of the water evaporates, it has a new mass of 29.2 g. What is the molarity of the original calcium carbonate solution?

Answer 1

Explanation:

The given data is as follows.

Mass of empty flask is 28.5 g.

Volume is 12.0 mL or $\frac{12mL \times 1 L}{1000 mL}$ = 0.012 L.

Mass of calcium carbonate and flask is 38.5 g.

Mass of evaporated solution in the flask is 29.2 g. This means that the flask now contains only calcium carbonate as all the water has been evaporated.

Therefore, mass of $CaCO_{3}$ = mass of evaporated solution in the flask - mass of empty flask

                                        = 29.2 g - 28.5 g

                                        = 0.7 g

As it is known that molar mass of $CaCO_{3}$ is 100 g/mol.

So, number of moles will be calculated as follows.

                     No. of moles = $\frac{mass}{molar mass}$

                                            = $\frac{0.7 g}{100 g/mol}$

                                            = 0.007 g

Since, molarity is the number of moles divided by volume in liter.

                       Molarity = $\frac{\text{no. of moles}}{\text{volume in liter}}$

                                      = $\frac{0.007 g}{0.012}$

                                      = 0.58 M

Thus, we can conclude that molarity of the original calcium carbonate solution is 0.58 M.