A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 N HCl is required. The amount of NaOH present in the original solution is :

A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 N H

1.	A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 N HCl is required. The amount of NaOH present in the original solution is :
Answer» <html><body><p>0.5 g<br/>1 g<br/>2 g<br/>4 g</p>Solution :300 mL HCl of 0.1 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> <a href="https://interviewquestions.tuteehub.com/tag/neutralises-2194775" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALISES">NEUTRALISES</a> entire amount of NaOH and `1//2 " of" Na_(2)CO_(3)`. Remaining `1//2 " of " Na_(2)CO_(3)` is neutralised by 25 mL 0.2 N HCl, i.e., 50 mL of 0.1 N HCl. <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, 250 mL of 0.1 N HCl is required to neutralise NaOH completely. <br/> `N_(1)V_(1)(NaOH)=N_(2)V_(2)(HCl)` <br/> `=0.1xx250` <br/> =25 <br/> `W_(NaOH)=(ENV)/(1000)=(40xx25)/(1000)=1 g`</body></html>