In Kjeldahl's method for estimation of nitrogen, $CuSO_4$ acts as

Question

5.33 gram of $CrCl_3 \cdot 6H_2O$ (1:3 electrolyte) is passed through cation exchanger. The resulting solution is then treated with an excess of $AgNO_3$, leading to formation of 8.61 gran of precipitate. Calculate :
$\frac{\text{number of moles of complex reacted}}{\text{number of moles of AgCl precipitated}} \times 100$

Answer 1

The correct option is (C): Catalytic agent.

Column-I (Complex compound)	Column-II ($\Delta_0$ (CFSE) $\text{cm}^{-1}$)
(i) $[Cr(CN)_6]^{3-}$	(P) 17000
(ii) $[Cr(H_2O)_6]^{3+}$	(Q) 15000
(iii) $[Cr(en)_3]^{3+}$	(R) 12000
(iv) $[CrF_6]^{3-}$	(S) 20000

In Kjeldahl's method for estimation of nitrogen, \(CuSO_4\) acts as

The Correct Option is C

Solution and Explanation

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