When `HNO_(3)` is converted into `NH_(3)`, the equivalent weight of `HNO_(3)` will be:

When N_(2) is converted into NH_(3) , the equivalent weight of nitrogen will be

When N_(2) is converted into NH_(3) , the equivalent weight of nitrogen will be:

When dil. HNO_(3) is electrolysed

When dil. HNO_(3) is electrolysed

Normally: An aquous solution contains 4.202 g of HNO_(3) in 600 mL of solution. Calculate the normally of solution? Stragegy: Convert grams of HNO_(3) to moles of HNO_(3) to moles of HNO_(3) and then to equivalents of HNO_(3) . Finally , apply the defintion of normally. (g HNO_(3))/(L) rarr (mol HNO_(3))/(L) rarr (eq HNO_(3))/(L) rarr (eq HNO_(3))/(L) = N HNO_(3)

One mole of As_(2)S_(3) is oxidized by HNO_(3) to H_(3)AsO_(4) and H_(2)SO_(4). HNO_(3) is converted into NO. The moles of HNO_(3) required are :

The equivalent weight of a species if acts as oxidant or reductant should be derived by : Eq. weight of oxidant or reductant = ("Mol. wt. of oxidant or reductant")/{("Number of electrons lost or gained by one"),("moleculae of oxidant or reductant"):} During chemical reactions, equal equivalents of one species react with same number of equivalents of other species giving same number of equivalent of products. However this is not true for reactants if they react in terms of moles. Also Molarity can be converted to normality by multiplying the molarity with valence factor or 'n' factor. One mole of As_(2)S_(3) is oxidised by HNO_(3) to H_(3)AsO_(4) and H_(2)SO_(4).HNO_(3) is converted into NO . The moles of HNO_(3) required are:

When Sn is treated with conc. HNO_(3)