$3.36 g$ of a dibromide of element X, $X B r_{2}$ is completely converted into dichloride $X C l_{2}$ forming $1.58 g$ of it $(X C l_{2})$
$X B r_{2} + C l_{2} \to X C l_{2} + B r_{2}$
What is the atomic mass of element X?

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Solution

Given,
$X B r_{2} \to X C l_{2}$
Let, atomic mass of $X$ be $Z$
Then, molar mass of $X B r_{2} = Z + 160 g$
molar mass of $X C l_{2} = Z + 71 g$

According to the question,
$\frac{3.36}{Z + 160} = \frac{1.58}{Z + 71}$
$\Rightarrow 1.58 (Z + 160) = 3.36 (Z + 71)$
$\Rightarrow Z = 8$

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3.36 g of a dibromide of element X, XBr2 is completely converted into dichloride XCl2 forming 1.58 g of it (XCl2) XBr2+Cl2→XCl2+Br2 What is the atomic mass of element X?

Given, XBr2→XCl2 Let, atomic mass of X be Z Then, molar mass of XBr2=Z+160 g molar mass of XCl2=Z+71 g According to the question, 3.36Z+160=1.58Z+71 ⇒1.58(Z+160)=3.36(Z+71) ⇒Z=8

$3.36 g$ of a dibromide of element X, $X B r_{2}$ is completely converted into dichloride $X C l_{2}$ forming $1.58 g$ of it $(X C l_{2})$
$X B r_{2} + C l_{2} \to X C l_{2} + B r_{2}$
What is the atomic mass of element X?

Given,
$X B r_{2} \to X C l_{2}$
Let, atomic mass of $X$ be $Z$
Then, molar mass of $X B r_{2} = Z + 160 g$
molar mass of $X C l_{2} = Z + 71 g$

According to the question,
$\frac{3.36}{Z + 160} = \frac{1.58}{Z + 71}$
$\Rightarrow 1.58 (Z + 160) = 3.36 (Z + 71)$
$\Rightarrow Z = 8$