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.

448 mL of SO_2 at NTP is passed through 100 mL. of a 0.2 N solution of

1.	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.
Answer» Solution :`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` willalso 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.` )