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JEE Main & Advanced Sample Paper JEE Main Sample Paper-1

  • question_answer
    3.7g of an oxide of a metal was heated with charcoal. The liberated \[C{{O}_{2}}\]was absorbed in caustic soda solution and weighed 1.0 g. If the specific gravity of the metal is 0.095, the exact atomic weight of the metal is

    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= 65.4


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