Question

The solubility of calcium carbonate is 14 . This rate means that 14 milligrams of calcium carbonate can dissolve in 1 liter of water. How much water would be required to fully dissolve 11 grams of calcium carbonate? Express your answer to the correct number of significant figures. One milligram is equal to 0.001 grams. It would take liters of water to fully dissolve 11 grams of calcium carbonate.

Answer 1

11 g CaCO3 * 1 L water / 14 mg CaCO3 * 1000 mg CaCO3 / 1 g CaCO3 = 790 L water.

Answer 2

Answer:- 786 L of water is required.

Solution:- From given information, solubility of calcium carbonate is
`14(mg)/(L)` .

We are asked to calculate the volume of water required to dissolve 11 grams of calcium carbonate. Solubility is like the density of the solution and the mass is also given. So, this problem is like to calculate the volume when mass and density are given.

Density formula is:

`d=(M)/(V)`

where d is density, M is mass and V is volume.

using this formula, we can calculate the volume but unit conversion is required as the mass is given in grams and the density gas milligrams. So, let's first convert 11 grams to milligrams. (1g = 1000mg)

`11g((1000mg)/(1g))`

= 11000 mg

The formula could be rearranged for the volume as:

`V=(M)/(d)`

Let's plug in the values in it:

`V=(11000mg)/((14mg)/(L))`

V = 785.71 L

It could be round to 786 L. So, 786 L of water is required to dissolve 11 grams of calcium carbonate. (Note- If significant figures are considered then it is round off to 790 L)