A) 170.8
B) 32.7
C) 67.37
D) 65.4
Correct Answer: D
Solution :
Weight of \[C{{O}_{2}}=1g\](as absorbed in KOH)Weight of oxygen in oxide = weight of oxygen in 1 g of \[C{{O}_{2}}=\frac{32}{44}=\frac{8}{11}g\] Weight of metal \[=3.7-\frac{8}{11}=32.7\] Equivalent wt. \[=\frac{\text{wt}\text{.}\,\text{of}\,\text{metal}}{\text{wt}\text{.}\,\text{of}\,\text{oxygen}}\times 8=32.7\] According to Dulong Petits law: Atomic weight (approx.)\[=\frac{6.4}{0.095}=67.37\] Valency \[=\frac{\,\text{atomic}\,\text{wt}\text{.}}{\text{equivalent}\,\text{wt}\text{.}}=2\,\text{aaprox}\text{.}\] Exact atomic weight = 32.7 x 2= 65.4You need to login to perform this action.
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