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Two elements A and B form compounds having molecular formula `AB_(2)` and `AB_(4)`. When dissolved in 20g of benzene, 1g of `AB_(2)` lowers the fpt by 2.3K whereas 1g of `AB_(4)` lowers it by `1.3K (K_(f)` for `C_(6)H_(6) = 5.1 K m^(-1))` At. Mass of A and B areA. `42.64` and `25.58`B. `25.58` and `42.64`C. `22.64` and `55.58`D. `45.64` and `22.58` |
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Answer» Correct Answer - B `2.3 = (5.1 xx 1 xx 1000)/((a+2b)xx 20)` `a +2b = (51 xx 50)/(23)` ..(i) Similary, `a +4b = (51 xx 50)/(13)` ...(ii) Subtracting (i) from (ii) `2b = (51 xx 50 xx 10)/(13 xx 23)` `b = 42.64` Substituting in (i). `a = 25.58` |
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