A metal oxide has the formula $X_{2} O_{3}$ .It can be reduced by hydrogen to give free metal and water.
0.1596 gr of metal oxide requires 6 mg of hydrogen for complete reduction.The atomic mass of metal in amu is

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55.8

5.58

155.8

15.58

answer is D.

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Detailed Solution

X_{2} O_{3} + 3 H_{2} \to 2 X + 3 H_{2} O 1 mol 3 mol 2 a + 48 6 gr ∵ 0.006 gr H_{2} is required by 0.1596 gr oxide ∴ 6 gr H_{2} will be required by 159.6 gr oxide ∴ 2 a + 48 = 159.6 a = 55.8 Where, a = atomic mass of metal' M'

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