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A 0.5 percent aqueous solution of KCl was found to freeze at 272.76K. Calculate the Van.t Hoff factor and degree of dissociation of the solute at this concentration (K_(f)" for water = 1.86 k.kg.mol"^(-1)). Normal molar mass of KCl = 74.5.

Answer» <html><body><p></p>Solution :`T^(@)<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>" of water "=273K` <br/> `T_(f)" of the solution "=272.76K` <br/> `DeltaT_(f)=T_(f^(@))-T_(f)=+0.24K` <br/> `M_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=(K_(f)W_(2))/(DeltaT_(f).W_(1))` <br/> Observed molecular mass <br/> `M_(2)=(1.86xx0.5xx1000)/(100xx0.24)="38.75 g.mol"^(-1)` <br/> The colligative property is <a href="https://interviewquestions.tuteehub.com/tag/inversely-7688775" style="font-weight:bold;" target="_blank" title="Click to know more about INVERSELY">INVERSELY</a> related to the molar mass. <br/> `therefore"Van.t Hoff factor"` <br/> `i=("Observed colligative property")/("<a href="https://interviewquestions.tuteehub.com/tag/normal-1123860" style="font-weight:bold;" target="_blank" title="Click to know more about NORMAL">NORMAL</a> colligative property")` <br/> `=("Theoretical molar mass")/("Observed molar mass")` <br/> `"Vant Hoff factor (i) "=(74.5)/(38.75)=1.92` <br/> `"Degree of dissociation "alpha=(i-1)/(n-1)` <br/> `n="2 for <a href="https://interviewquestions.tuteehub.com/tag/kcl-527837" style="font-weight:bold;" target="_blank" title="Click to know more about KCL">KCL</a>"` <br/> `therefore alpha=(1.92-1)/(2-1)` <br/> `therefore"Degree of dissociation "=0.92`</body></html>


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