Question

How many molecules of CaCl2 are equivalent to 75.9 g CaCl2

Answer 1

Answer: 4.11 * 10^ 23 molecules

Step-by-step 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)` =
`(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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