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1.325 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 250 mL. On titration 25 mL of this solution neutralise 20 mL of a solution of suphuric acid. How much water should be added to 450 mL of this acid solution to make it exactly N//12 ?

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. mass of `Na_(2)CO_(3)=("Mol.mass")/(2)=(106)/(2)=<a href="https://interviewquestions.tuteehub.com/tag/53-324827" style="font-weight:bold;" target="_blank" title="Click to know more about 53">53</a>` <br/> 250 mL of the <a href="https://interviewquestions.tuteehub.com/tag/sodium-1215510" style="font-weight:bold;" target="_blank" title="Click to know more about SODIUM">SODIUM</a> carbonate solution contains = 1.325 g <br/> 1000 mL of the sodium carbonate solution contains <br/> `=(1.325g)/(250)xx1000=5.300` g<br/> <a href="https://interviewquestions.tuteehub.com/tag/normality-1124423" style="font-weight:bold;" target="_blank" title="Click to know more about NORMALITY">NORMALITY</a> of `Na_(2)CO_(3)` solution` =("Strength "(g//L))/("Eq. mass")` <br/> `=(5.30)/(53)=(1)/(10)N` <br/> Applying `<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>((Na_(2)CO_(3)))(N_(1)V_(1))-=underset((H_(2)SO_(4)))(N_(2)V_(2))` <br/> `(1)/(10)xx25=N_(2)xx20` <br/> `N_(2)=(25)/(10xx20)=(1)/(8)` <br/> Applying`underset(("Before dilution"))(N_(B)V_(B))-=underset(("After dilution"))(N_(A)V_(A))` <br/> `=(1)/(8)xx450=(1)/(12)xxV_(A)` <br/> `V_(A)=(450xx12)/(8)=675 mL` <br/> Water to be added for dilution =(675-450)=225 mL</body></html>


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