JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    A transition metal M forms a volatile chloride which has a vapour density of 94.8. If it contains74.75% of chlorine the formula of the metal chloride will be .   JEE Main Online Paper (Held On 26-May-2012)     

    A) \[MC{{l}_{3}}\]                 

    B) \[MC{{l}_{2}}\]

    C) \[MC{{l}_{4}}\]                 

    D) \[MC{{l}_{5}}\]

    Correct Answer: C

    Solution :

    74.75% of chlorine means 74.75g chlorine is present in 100g of metal chloride. Weight of metal =\[=100g-74.75g\] \[=25.25g\] Equivalent weight\[\text{=}\frac{\text{weight}\,\text{of}\,\text{metal}}{\text{weight}\,\text{of}\,\text{chlorine}}\times 35.5\] \[=\frac{25.25}{74.75}\times 35.5=12\]Velency of metal \[\text{=}\frac{\text{2 }\!\!\times\!\!\text{ V}\text{.D}\text{.}}{\text{Equivalent}\,\text{wt}\text{.}\,\text{of}\,\text{metal}\,\text{+35}\text{.5}}\] \[\text{=}\frac{\text{2 }\!\!\times\!\!\text{ 94}\text{.8}}{12+35.5}=4\] \[\therefore \]Formula of compound \[=MC{{l}_{4}}\]


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