A) 23.9
B) 27.3
C) 36.3
D) 48.3
Correct Answer: A
Solution :
Given equivalent weight of metal = 9 Vapour density of metal chloride = 59.25 \[\therefore \] molecular weight of metal chloride \[=2\times V.D=2\times 59.25=118.5\] \[\therefore \] valency of metal \[=\frac{\text{molecular weight of metal chloride}}{\text{equivalnet weight of metal }+\text{35}\text{.5}}\] Valency of metal \[=\frac{118.5}{9+35.5}=\frac{118.5}{44.5}=2.66\] Therefore atomic weight of the metal =equivalent weight \[\times \] valency \[=9\times 2.66=23.9\]
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