How many grams of CaCl2 would be dissolved in 1.4 L of a 0.11 M solution of CaCl2?
mols = M x L = ?
Then mols = grams/molar mass
You know mols and grams, solve for grams.
To find out how many grams of CaCl2 would be dissolved in 1.4 L of a 0.11 M solution of CaCl2, we can use the formula:
Amount of solute (in moles) = concentration (in M) x volume (in liters)
First, convert the given concentration of 0.11 M to moles per liter (mol/L).
0.11 M means there are 0.11 moles of CaCl2 dissolved in 1 liter of solution.
Next, multiply the concentration (0.11 mol/L) by the volume of the solution (1.4 L) to find the number of moles:
0.11 mol/L x 1.4 L = 0.154 moles of CaCl2
Now, we need to convert moles to grams using the molar mass of CaCl2.
The molar mass of CaCl2 is equal to the sum of the atomic masses of calcium (Ca) and two chlorine (Cl) atoms.
Ca: 40.08 g/mol
Cl: 35.45 g/mol
Mass of CaCl2 = (40.08 g/mol of Ca + 2 * 35.45 g/mol of Cl) = 110.98 g/mol
Finally, multiply the moles of CaCl2 by the molar mass to find the grams of CaCl2 dissolved:
0.154 moles x 110.98 g/mol = 17.12 grams
Therefore, approximately 17.12 grams of CaCl2 would be dissolved in 1.4 L of a 0.11 M solution of CaCl2.