Question

A certain compound has the molecular formula X406 having 57.2%. X thus:
(A) 'X' can be non metal
(B) 'X' is an electro positive metal
(C) atomic mass of X is 32
(D) 'X' may contains 3 valence electrons​

Answer 1

Answer:

Mass of the Compound (X₄O₆) = 10 g.

Mass of the Element X in the Compound = 6.06 g.

Let the Atomic Mass of the Element X be a.

∴ Mass of X in the compound = 4a.

∴ Molecular Mass of the Compound = a × 4 + 16 × 6

   = 4a + 96

  

Now,

Mass of X/Mass of the compound = At. Mass of the X/Mol. Mass of the Compound.

⇒ 6.06/10 = 4a/(4a + 96)

⇒ 6.06/10 = 4a/4(a + 24)

⇒ 0.606 = a/(a + 24)

⇒ 0.606(a + 24) = a

⇒ 0.606a + 14.544 = a

⇒ a - 0.606a = 14.544

⇒ 0.394a = 14.544

⇒ a = 14.544/0.394

∴ a = 36.9

∴ a ≈ 37 a.m.u.

Hence, the Atomic Mass of the Element X is 37 a.m.u.

Hope it helps.

Explanation: