The reaction of H₃N₃B₃Cl₃ (A) with LiBH₄ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H₃N₃B₃(Me)₃. Compounds (B) and (C) respectively, are :

A, B and C are three biomolecules. The results of the tests performed on them are given below:

$$ \begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Molisch's } \\ \text { Test } \end{array} & \begin{array}{l} \text { Barfoed } \\ \text { Test } \end{array} & \begin{array}{l} \text { Biuret } \\ \text { Test } \end{array} \\ \hline \text { A } & \text { Positive } & \text { Negative } & \text { Negative } \\ \hline \text { B } & \text { Positive } & \text { Positive } & \text { Negative } \\ \hline \text { C } & \text { Negative } & \text { Negative } & \text { Positive } \\ \hline \end{array} $$

A, B and C are respectively :

5 g of zinc is treated separately with an excess of

(a) dilute hydrochloric acid and
(b) aqueous sodium hydroxide.

The ratio of the volumes of H₂ evolved in these two reactions is :

The solubility product of Cr(OH)₃ at 298 K is 6.0 × 10^–31. The concentration of hydroxide ions in a saturated solution of Cr(OH)₃ will be :