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20 " mL of " `H_(2)O_(2)` is reacted completely with acidified `K_(2)Cr_(2)O_(7)` solution 40 " mL of " `K_(2)Cr_(3)O_(7)` solution was required to oxidised the `H_(2)O_(2)` completely. Also, 2.0 " mL of " the same `K_(2)Cr_(2)O_(7)` solution required 5.0 " mL of " a 1.0 M `H_(2)C_(2)O_(4)` solution to reach equivalence point. Which of the following statements is/are correct?A. The `H_(2)O_(2)` solution is 5 M.B. The volume strength of `H_(2)O_(2)` is 56V.C. The volume strength of `H_(2)O_(2)` is 112V.D. If 40 " mL of " `(5M)/(8H_(2)O_(2))` is further added to the 10 " mL of " above `H_(2)O_(2)` solution the volume strength of the resulting solution is changed to 16.8 V. |
Answer» Correct Answer - A::B::D (i). `mEq H_(2)O_(2)-=m" Eq of "Cr_(2)O_(7)^(2-)` `(n=2)(n=6)` `20mLxxN_(1)=40mL xxN_(2)` (a). `m" Eq of "Cr_(2)O_(7)^(2-)-=m" Eq of "C_(2)O_(4)^(2-)(n=2)` `2.0xxN_(2)-=5.0xx1.0xx2` `N_(2)(Cr_(2)O_(7)^(2-))=5` ..(ii) Therefore substituting the `N_(2)` in equation (i) `20 mLxxN_(1)=40mLxx5` `N_(1)(H_(2)O_(2))=10` `M_(1)(H_(2)O_(2))=(10)/(2)=5M` (b). `1 N H_(2)O_(2)=5.6V` `therefore10 N H_(2)O_(2)=56V` (d). `N_(1)V_(1)+N_(2)V_(2)=N_(3)V_(3)(V_(3)=10+40=50mL)` `10xx10+40xx(5)/(8)xx2=N_(3)xx50` `N_(3)` (final) `H_(2)O_(2)=3N` The volume strength of `H_(2)O_(2)=5.6xx3=16.8V` |
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