• # question_answer Calculate the mass of a non-volatile solute (molar mass $40\,g\,mo{{l}^{-\,1}}),$ which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. A) 20 g      B) 10 g     C) 30 g                              D) 45 g

[b]  According to Raoul?s law, relative lowering of vapour pressure, $\frac{p_{A}^{{}^\circ }-ps}{p_{A}^{{}^\circ }}={{\chi }_{B}}$ ${{\chi }_{B}}=\frac{{{n}_{B}}}{{{n}_{B}}+{{n}_{A}}}=\frac{{{W}_{B}}/{{M}_{B}}}{\frac{{{W}_{B}}}{{{M}_{B}}}+\frac{{{W}_{A}}}{{{M}_{A}}}}$ Given, vapour pressure is reduced to 80 when non-volatile solute is dissolved in octane. It means If $p_{A}^{{}^\circ }=1\,\text{atm}$then ${{p}_{S}}=0.8\,\text{atm};$ $p_{A}^{{}^\circ }-{{p}_{S}}=0.2\,\text{atm};$ ${{M}_{A}}({{C}_{8}}{{H}_{18}})=114\,g\,\text{mo}{{\text{l}}^{-\,1}};\,\,{{W}_{A}}=114g;$ ${{M}_{B}}=40g\,\text{mo}{{\text{l}}^{-\,1}};\,\,{{W}_{B}}=?$ ${{M}_{B}}=40g\,mo{{l}^{-1}};\,{{W}_{B}}=?$ On applying Eq. (i) and (ii), we get, $\frac{0.2}{1}=\frac{{{W}_{B}}40}{\frac{{{W}_{B}}}{40}+\frac{114}{114}}=\frac{{{W}_{B}}/40}{\frac{{{W}_{B}}}{40}+1}$ $0.2=\frac{{{W}_{B}}}{{{W}_{B}}+40}$ $0.2{{W}_{B}}+8={{W}_{B}}$ ${{W}_{B}}=10g$