• # 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 K.kg. $mo{{l}^{-1}}$, 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)  ${{1.5}^{o}}$

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