JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    Two elements 'A' and 'B' form three covalent non-volatile solids with molecular formula \[A{{B}_{2}}\],\[A{{B}_{4}}\] and\[A{{B}_{8}}\]. One gram of each is dissolved in 20 g \[{{C}_{8}}{{H}_{6}}(\ell )\] separately For the solutions of \[A{{B}_{2}}\] and \[A{{B}_{4}}\] lowering in freezing points are found to \[\frac{5}{2}\] and \[\frac{5}{3}\] degree respectively. Assuming that here is no dissociation or association of any compound and \[{{K}_{f}}\] for \[{{C}_{6}}{{H}_{6}}(\ell )\] be 5 \[mo{{l}^{-1}}\], what will be \[\Delta T\], for solution of \[A{{B}_{8}}\].

    A)  \[\frac{{{5}^{o}}}{4}\]                                  

    B)  \[\frac{{{8}^{o}}}{5}\]

    C)  \[{{1}^{o}}\]                                     

    D)  \[{{1.5}^{o}}\]

    Correct Answer: C

    Solution :

    Let atomic mass of A and B be \[{{M}_{A}}\And {{M}_{B}}\] respectively. For compound \[A{{B}_{2}}\]     \[\Delta {{T}_{f}}=i{{K}_{f}}m\] \[\frac{5}{2}\,=1(5)\,\frac{1/{{M}_{a}}\,+2{{M}_{B}}}{20/1000}\] \[100={{M}_{A}}+2{{M}_{B}}\]                     ?(i)  For compound \[A{{B}_{4}}\]     \[\Delta {{T}_{f}}=i{{K}_{f}}m\] \[\frac{5}{3}\,=1(5)\frac{1/{{M}_{a}}+4{{M}_{B}}}{20/1000}\] \[150\,={{M}_{A}}\,+4{{M}_{B}}\]     ?(ii) From equ.(i) &equ.(ii)      \[{{M}_{A}}=50\] \[{{M}_{B}}=25\] For compound \[A{{B}_{B}}\]\[\Delta {{T}_{f}}=i{{K}_{f}}m\] \[=1(5)\,\frac{1/{{M}_{a}}+8{{M}_{B}}}{20/1000}\]= 1

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