Question

calculate the number of atoms of chlorine in 2.08gm of Bacl2​

Answer 1

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Answer 2

Answer:

1.20 × $10^{22}$ number of atoms of chlorine is present in 2.08 gm of $BaCl_2$.

Explanation:

Weight of the mixture $BaCl_2$ = 2.08 gm

Atomic weight of Barium, Ba = 137 gm

Atomic weight of chlorine, cl = 35.5 gm

Thus, molar mass of $BaCl_2$ = 137 + 35.5 + 35.5

                                             = 208 gm/mol

Out of 208 grams of $BaCl_2$, 71 grams is chlorine.

Using unitary method,

⇒ $\frac{208}{100}$ grams of $BaCl_2$ contains $\frac{71}{100}$ grams of chlorine.

⇒ 2.08 grams of $BaCl_2$ contains 0.71 grams of chlorine.

Since 35.5 gm of chlorine is equivalent to 6.022 × $10^{23}$ atoms.

⇒ 1 gm of chlorine is equivalent to 6.022 × $\frac{10^{23}}{35.5}$ atoms.

⇒ 0.71 gm of chlorine is equivalent to 6.022 × $\frac{10^{23}}{35.5}$ × 0.71 atoms.

= 1.20 × $10^{22}$ atoms

Hence 1.20 × $10^{22}$ number of atoms of chlorine is present in 2.08 gm of $BaCl_2$.

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