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A metal oxide has the formula Z₂O_3.It can be reduced by hydrogen to give a free metal and water. 0.1596 gm of the metal oxide requires 6mg of hydrogen for complete reduction. The atomic mass of the metal is

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27.9

79.8

159.6

55.8

answer is D.

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Detailed Solution

Z₂O₃ + 3H₂ → 2Z + 3H₂O
Weight of Z₂O₃ = 0.1596g
Weight of Hydrogen = 6 x 10^-3 g

\large \\\\\frac{{{W_{{Z_2}{O_3}}}}}{{{W_H}}}\; = \;\frac{{{E_{{Z_2}{O_3}}}}}{{{E_H}}}\\\\

\large \\\\\frac{{{W_{{Z_2}{O_3}}}}}{{{W_H}}}\; = \;\frac{E_Z+E_{O^{-2}}}{{{E_H}}}\\\\

\large \frac{{0.1596}}{{6 \times {{10}^{ - 3}}}}\; = \;\frac{{{E_Z}\; + \;8}}{1}

\large \frac{{159.6}}{6}\; = \;{E_Z}\; + \;8

E_Z = 18.6
Ew of metal = 18.6
Formula of metal Oxide = Z₂O₃
Valency of 'Z' = 3;Aw of metal = (18.6 x 3) = 56

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