Answer
To determine the mass of NaCl that would contain the same total number of ions as 284 g of MgCl2, we need to use Avogadro's number and the molar mass of the compounds.
Work Step by Step
Step 1: Determine the total number of ions in 284 g of MgCl2.
The molar mass of MgCl2 is 95.21 g/mol. Therefore, the number of moles of MgCl2 can be calculated as:
moles of MgCl2 = mass / molar mass = 284 g / 95.21 g/mol = 2.985 mol
Each molecule of MgCl2 dissociates into three ions (one Mg2+ ion and two Cl- ions). Therefore, the total number of ions in 284 g of MgCl2 is:
total number of ions = 2.985 mol x 3 ions/mol = 8.955 x 10^23 ions
Step 2: Determine the number of ions per unit mass for MgCl2.
The number of ions per unit mass can be calculated as:
number of ions per unit mass = total number of ions / mass = 8.955 x 10^23 ions / 284 g = 3.15 x 10^21 ions/g
Step 3: Calculate the mass of NaCl that would contain the same number of ions.
The molar mass of NaCl is 58.44 g/mol. Therefore, the mass of NaCl that would contain the same number of ions as 284 g of MgCl2 can be calculated as:
mass of NaCl = (total number of ions of MgCl2) x (molar mass of NaCl) / (number of ions per unit mass of MgCl2)
Substituting the values from previous steps, we get:
mass of NaCl = (8.955 x 10^23 ions) x (58.44 g/mol) / (3.15 x 10^21 ions/g) = 166 g
Therefore, 166 g of NaCl would contain the same total number of ions as 284 g of MgCl2.
