`448` mL of `SO_2` at NTP is passed through 100 mL. of a 0.2 N solution of `NaOH`. Find the weight of the salt formed.

Explanation

`SO_2 + NaOH to NaHSO_3`
m.e. of `NaOH = 0.2 xx 100 = 20. " ...(Eqn. 1, Chapter 7)"`
Equivalent of NaOH `= ( 20)/( 1000) = 0.02 " . .. (Eqn. 3, Chapter 7)"`
Since 1 equivalent of NaOH combines with 1 mole of `SO_2` according to the above reaction,
`therefore` for `SO_2`, 1 mole = 1 equivalent, i.e., 1 equivalent of `SO_2` will occupy 22.4 litres at NTP.
Equivalent of `SO_2 = ( 448)/( 22400)`
`= 0.02`
We see that number of equivalents of `SO_2` and that of NaOH are equal. Number of eq. of ` NaHSO_3` will also be 0.02 by the law of equivalence.
`therefore` weight of `NaHSO_3` = equivalent of `SO_2xx` equivalent weight
`=0.02 xx 104`
`=2.08`g.
(According to the given reaction, eq. wt. of `NaHSO_3=` mol. wt. `=104.` )