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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» <html><body><p></p>Solution :`SO_2 + <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> to NaHSO_3` <br/> m.e. of `NaOH = 0.2 xx 100 = 20. "...(Eqn. 1, Chapter 7)"` <br/> <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> of NaOH `= ( 20)/( <a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>) = 0.02 ". .. (Eqn. 3, Chapter 7)"` <br/> Since 1 equivalent of NaOH combines with 1 mole of `SO_2` according to the above <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a>, <br/> `therefore` for `SO_2`, 1 mole = 1 equivalent, i.e., 1 equivalent of `SO_2` will occupy 22.4 litres at NTP. <br/> Equivalent of `SO_2 = ( 448)/( 22400)` <br/> `= 0.02` <br/> 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 <a href="https://interviewquestions.tuteehub.com/tag/equivalence-452629" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENCE">EQUIVALENCE</a>. <br/>`therefore` weight of `NaHSO_3` = equivalent of `SO_2xx` equivalent weight <br/> `=0.02 xx 104` <br/> `=2.08`g. <br/> (According to the given reaction, eq. wt. of `NaHSO_3=` mol. wt. `=104.` )</body></html>


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