Question

Calculate the number of atoms of chlorine in 2.08
gm of BaCl2​

Answer 1

Answer:

Chlorine atomic mass is 35.5g.

Atomic mass of Barium is 56g.

Therefore, molar mass of BaCl2 is 127g.

Of 127g BaCl2, 71g is Chlorine.

Using unitary method, of 2.08g BaCl2, 1.163g is Chlorine.

35.5g of Chlorine is equivalent to 6.022X10^23 atoms.

Thus, 1.163g chlorine is equivalent to 1.97X10^22 atoms.

Answer 2

The mixture's ($BaCl_{2}$) weight is 2.08 grams.

Barium, Ba has an atomic weight of 137 grams and Chlorine, Cl has an atomic weight of 35.5 grams.

As a result, $BaCl_{2}$ 's molar mass is 137 + 35.5 + 35.5 = 208 gm/mol.

71 grams of the $BaCl_{2}$ are of chlorine.

Using the unitary approach, $\frac{71}{100}$ grams of chlorine are found in $\frac{208}{100}$ grams of $BaCl_{2}$.

2.08 grams of $BaCl_{2}$ have 0.71 grams of chlorine in them.

1 gram of chlorine is equivalent to $6.022$ × $\frac{10^{23} }{35.5}$ atoms since 35.5 gram of chlorine is equal to $6.022$ × $10^{23}$ atoms.

0.71 grams of chlorine are equal to $6.022$ × $\frac{10^{23} }{35.5}$ × $0.71$ atoms, or $1.20$ ×$10^{22}$ atoms.

Consequently, $1.20$ × $10^{22}$ chlorine atoms are contained in 2.08 g of $BaCl_{2}$.

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