0.5 g KCl was dissolved in 100 g water and the solution originally at 20^(@)C, froze at -0.24^(@)C. Calculate the percentage ionization of salt. K_(f) per 1000 g of water =1.86^(@)C.

0.5 g KCl was dissolved in 100 g water and the solution originally at

1.	0.5 g KCl was dissolved in 100 g water and the solution originally at 20^(@)C, froze at -0.24^(@)C. Calculate the percentage ionization of salt. K_(f) per 1000 g of water =1.86^(@)C.
Answer» Solution :Observed mol.Mass is obtained from the given data, i.e., `w_(2)="0.5 g,"w_(1)="100 g,"M_(1)="18 g mol"^(-1)" for "H_(2)O, DeltaT_(f)=0-(-0.24)=0.24^(@)C` `M_(2)=(1000K_(f)w_(2))/(DeltaT_(f)xxw_(1))=("1000 g kg"^(-1)xx"1.86 K kg mol"^(-1)xx"0.5 g")/("0.24 K"xx"100 g")="38.75 g mol"^(-1)` Calculated (theoretical) mol mass of KCl `=39+35.5="74.5 g mol"^(-1)` `therefore"i"=("Calculated mol mass")/("Observed mol mass")=(74.5)/(38.75)=1.92` Now, KCl dissociated as `{:(,KCl,hArr,K^(+),+,CL^(-)),("Initial moles","1 mole",,0,,0),("Moles after disso.",1-alpha,,alpha,,alpha):}` `"Total no. of moles after dissociation "=1-alpha+alpha+alpha+=1+alpha""therefore""i=(1+alpha)/(1) or alpha=i-1=1.92 -1=0.92` PERCENTAGE ionization `=0.92xx100=92%`.