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8 g of NaOH is dissolved in 18 g of H2O Mole fraction of NaOH in solution and molality (in mol kg-1) of the solution respectively are
1. 0.2, 11.11
2. 0.167, 22.20
3. 0.2, 22.20
4. 0.167, 11.11

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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.

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