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QUESTION
Q36) Vapour density of metal chloride is 66. Its oxide contains 53% metal. The atomic weight of the metal is ... [ Bihar MADT 1982]




SOLUTION

Equivalent Wt. of metal = Equivalent Wt. of oxygen
53/E = 48/8
E = 9.02
Let A be the atomic mass of metal
Then valency of Metal be (n) = A/E = A/9.02 ----(i)
Let Chloride be MCln
Now Molecular weight of Metal chloride=2 × Vapour density
Moleculat weight of metal chloride=2×66=132
Molecular wt of chloride = (A) + n(35.5)
From (i)
132 = (A) + (A/9.02) 35.5
132 = A[1 +(1/9.02×35.5]
On simplifying A = 26.72 say 27
Short method
47 is oxygen and 53 is metal
Number of oxygen atoms=47/16=3 implies that Valency of Metal=3
Now Molecular weight of Metal chloride=2 × Vapour density
Moleculat weight of metal chloride=2×66=132
Now 132=M + 3×35.5
M=132 - 3×35.5=25.5 say 27
Answer: (c)