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Q.

A metal oxide has the formula Z2O3. 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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a

27.9

b

79.8

c

159.6

d

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


EZ = 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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