1.

25 ml of FeC_2O_4 dissolved in 186 gm of water calculate depression in freezing point. It 10 ml of same FeC_2O_4 titrated with 30 ml of 0.4 M KMnO_4 in acidic medium (k_f for H_2O=1.86,Assume 100% ionisation of FeC_2O_4).

Answer»


Solution :`x_(FeC_2O_4)=x_(KMnO_4) " " undersetn(FeC_2O_4)toundersetn(Fe^(+2))+undersetn(C_2O_4^(-2))`
`x_(Fe^(+2))+x_(C_2O_4^(-2))=x_(KMnO_4) " " n_(FeC_2O_4)=(MV)/1000`
`1xxn+2xxn=(0.4xx5xx30)/1000 " " 0.02=(Mxx10)/1000`
`n=(2xx10)/1000=0.02 " " M=2`
n=no . of mole of SUBSTANCE
X=equivalent of substance
`DeltaT_f=i K_fm=2xx1.86xx((25xx2)/1000xx1000/186)=100/100=1`


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