Answer:
The identity of unknown substance is Barium.
Explanation:
The given compound has a formula of MCl2. And the percentage of chlorine in this compound is said to be 34.1 %. But, we know that the mass of chlorine in this compound is, 35.5*2 = 71 g
Therefore, if we let x be the total mass of the compound, then:
34.1% of x = 71 g
(0.341)x = 71 g
x = 71 g/0.341
x = 208.2 g
Hence, the mass of other atom M, must be:
M = x - 71 g
M = 208.2 g - 71 g
M = 137.2 g
Now, we look into periodic table in group II. We find that the element is Barium with atomic mass of 137 g.
The identity of unknown substance is Barium.
Barium forms a metal chloride whose percentage of chlorine is 34.1 %. The unknown substance is barium chloride.
We have a Group II metal chloride with a general formula MCl₂. The molar mass of Cl⁻ is 35.45 g/mol. The mass of 2 moles of Cl⁻ is:
[tex]2 mol Cl^{-} \times \frac{35.45g Cl^{-} }{1mol Cl^{-} } = 70.90g Cl^{-}[/tex]
70.90 g represents 34.1% of 1 mole of MCl₂. The molar mass of MCl₂ is:
[tex]70.90gCl^{-} \times \frac{100gMgCl_2}{34.1gCl^{-} } = 208gMgCl_2[/tex]
208 g of MCl₂ contains 70.90 g of Cl⁻. The mass of M²⁺ in 1 mole of MCl₂ is:
[tex]mMCl_2 = mCl^{-} + mM^{2+} \\\\mM^{2+} = mMCl_2 - mCl^{-} = 208g-70.9 g = 137.1 g[/tex]
Barium has a molar mass of 137.3 g/mol, so M must be barium. The unknown substance is barium chloride.
Barium forms a metal chloride whose percentage of chlorine is 34.1 %. The unknown substance is barium chloride.
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