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 `KMn

Question

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 `KMn

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`).

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - 1
`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`