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`4.5gL^(-1)``4.9gL^(-1)``2.25gL^(-1)``2.45gL^(-1)`

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In first titration, both `H_(2)SO_(4)` and `H_(2)C_(2)O_(4)` react with base, as acid-base titration n factor for both is 2 due to `2H^(o+)` per mol in each case. Let `x=m" Eq of "H_(2)C_(2)O_(4)` and `y=m" Eq of "H_(2)SO_(4)` <br> `thereforem" Eq of "H_(2)C_(2)O_(4)+m" Eq of "H_(2)SO_(4)=m" Eq of "KOH` <br> `thereforex+y=20xx0.1xx1` (n-factor) <br> `x+y=2` mEq ..(ii) <br> In seconds titration only `H_(2)C_(2)O_(4)` (being reducing agent) <br> Reacts with `K_(2)C_(2)O_(7)`. <br> `m" Eq of "H_(2)C_(2)O_(4)-=m" Eq of "K_(2)Cr_(2)O_(7)` <br> `(C_(2)O_(4)^(2-)to2CO_(2)+2e^(-))(6e^(-)+Cr_(2)O_(7)^(2-)to2Cr^(3+))` <br> `(x)-=m" Eq of "H_(2)C_(2)O_(4)-=50xx(1)/(300)xx6` <br> `thereforex-=1mEq` ..(ii) <br> From equations (i) and (ii) we get <br> `x=1 mEq, y=1 mEq` <br> Weight of `H_(2)C_(2)O_(4)=1xx10^(-3)xx45` <br> `(Ew of H_(2)C_(2)O_(4)=(90)/(2)=45)` <br> `=(0.045g)/(10mL)` <br> `=(0.045xx1000)/(10)gL^(-1)=4.5gL^(-1)` <br> Therefore, strength of `H_(2)C_(2)O_(4)=4.5gL^(-1)`