Question

How many molecules of CaCl2 are equivalent to 75.9 g CaCl2

Answer 1

Answer: 4.11 * 10^ 23 molecules

Explanation:

1) Data:

Compound: CaCl2

mass, m = 75.9 g

unknown, N = ?

2) Formulas:

N = n * Avogadro's number = n * 6.022 * 10^23 molecules / mol

n = mass in grams / molar mass

molar mass = sum of the masses of the atoms in the molecular formula

3) Solution

molar mass of CaCl2:

      mass of Ca = 1 * atomic mass of Ca = 40.1 g/mol
      mass of Cl2 = 2 * atomic mass of Cl2 = 2 * 35.5 g/mol = 71.0 g/mol
      molar mass = 40.1 g/mol + 71.0 g/mol = 111.1 g/mol

n = m / molar mass = 75.9 g / 111.1 g/mol = 0.683 mol

N = n * 6.022 * 10^23 = 0.683 mol * 6.022 * 10^23 molecules / mol = 4.11 * 10^23 molecules, which is the answer.

Answer 2

First, you need to find:
One mole of $CaCl_{2}$ is equivalent to how many grams?

Well, for this you have to look up the periodic table. According to the periodic table:
The atomic mass of Calcium Ca = 40.078 g (See in group 2)
The atomic mass of Chlorine Cl = 35.45 g (See in group 17)

As there are two atoms of Chlorine present in $CaCl_{2}$, therefore, the atomic mass of $CaCl_{2}$ would be:

Atomic mass of $CaCl_{2}$  = Atomic mass of Ca + 2 * Atomic mass of Cl

Atomic mass of $CaCl_{2}$ = 40.078 + 2 * 35.45 = 110.978 g

Now,

110.978 g of $CaCl_{2}$ = 1 mole.
75.9 g of $CaCl_{2}$ = $\frac{75.9}{110.978} moles$ = 0.6839 moles.

Hence,
The total number of moles in 75.9g of $CaCl_{2}$ = 0.6839 moles

According to Avogadro's number,
1 mole = 1 * $6.022 * 10^{23}$ molecules
0.6839 moles = 0.6839 * $6.022 * 10^{23}$ molecules = $4.118*10^{23}$ molecules

Ans: Number of molecules in 75.9g of  $CaCl_{2}$ =  $4.118*10^{23}$ molecules

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