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)`

Explanation

Step 1: Calculate the number of moles.
mol `HNO_(3) = ("Grams" HNO_(3))/("Molar mass" HNO_(3)) = (4.202g)/(63.02g mol^(-1))`
`= 0.0666 mol HNO_(3)`
Step 2: Calculate the number of equivalnets of `HNO_(3)`.
Number of equivalnents of `HNO_(3)` (No. eq `HNO_(3)`)
= (Number of moles of acid)
`xx` (Number of hydrogen ions furnished)
`= (0.0666 mol HNO_(3))(1)`
`= 0.0666 eq HNO_(3)`
Step 3: Calcularte normally.
Normally `(N) = ("No. eq" HNO_(3))?("Liners of solution")`
`= (0.0666 eq HNO_(3))/(0.600 L)`
`= 0.11 NHNO_(3)`
Alterntively using single setup,
`? (eq HNO_(3))/(L) = (4.202 g HN O_(3))/(0.600 L) xx (1 mol HNO_(3))/(63.02g HNO_(3))`
`xx (1 eq HNO_(3))/(1 mol HNO_(3))`
`= 0.0111 N NHO_(3)`