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In the standardization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by iodometry, th equivalent weight of `K_(2)Cr_(2)O` isA. `("Molecular weight")//2`B. `("Molecular weight")//6`C. `("Molecular weight")//3`D. Same as molecular weight |
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Answer» Correct Answer - B In iodometry, `H_(2)Cr_(2)O_(7)` liberates `I_(2)` from iodides (Nal or Kl) which is titrated with `Na_(2)S_(2)O_(3)` solution `K_(2)Cr_(2)O_(7) + I^(-) + H^(+) rarr Cr^(3+) + I_(2)` Here, one mole of `K_(2)Cr_(2)O_(7)` accepts 6 mole of electrons. `:.` Equivalent weight `= ("Molecular weight")/(6)` |
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