Correct Answer - Option 4 : 0.167, 11.11
Calculation:
Mole fraction of solute \(= {\rm{\;}}\frac{{{\rm{number\;of\;moles\;of\;solute\;}} + {\rm{\;number\;of\;moles\;solvent}}}}{{{\rm{number\;of\;moles\;of\;solute}}}}\)
\({\chi _{{\rm{solute\;}}}} = \frac{{{n_{{\rm{Solute\;}}}}}}{{{n_{{\rm{Solute\;}}}} + {n_{{\rm{Solvent\;}}}}}} = \frac{{\frac{{{{\rm{w}}_{{\rm{Solute}}}}{\rm{\;}}}}{{{\rm{M}}{{\rm{w}}_{{\rm{solute}}}}}}}}{{\frac{{{w_{{\rm{Solute\;}}}}}}{{M{w_{{\rm{Solute\;}}}}}} + \frac{{{\rm{\;}}{{\rm{w}}_{{\rm{solvent}}}}{\rm{\;}}}}{{{\rm{\;}}{{\rm{M}}_{\rm{w}}}_{{\rm{solvent}}}{\rm{\;}}}}}}\)
Given, wSolute = wNaoH = 8g
Mwsolute = MωNaOH = 40gmol-1)
wSolvent = vw(H2O) = 18g
MwSolvent = 18 gmol-1
\(\therefore {x_{{\rm{Solute\;}}}} = {\chi _{{\rm{NaOH\;}}}} = \frac{{\frac{8}{{40}}}}{{\frac{8}{{40}} + \frac{{18}}{{18}}}}\)
\(= \frac{{0.2}}{{0.2 + 1}} = \frac{{0.2}}{{1.2}} = 0.167\)
Now, molality \({\rm{\;}}\left( m \right) = \frac{{{\rm{\;Moles\;of\;solute\;}}}}{{{\rm{\;Mass\;of\;solvent\;}}\left( {{\rm{in\;kg}}} \right){\rm{\;}}}}\)
\(\frac{{\frac{{{w_{solute}}}}{{{M_w}solute}}}}{{{w_{solvent}}\left( {in\;g} \right)}} \times 1000\)
\(= \frac{{\frac{8}{{40}}}}{{18}} \times 1000 = \frac{{0.2}}{{18}} \times 1000 = 11.11{\rm{\;mol\;k}}{{\rm{g}}^{ - 1}}\)
Thus,
mole fraction of NaOH in solution and molality of the solution respectively are 0.167 and 11.11 mol kg-1.