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20 mL of a solution of H_2SO_4 neutralises 21.2 mL of 30% solution (w/v) of Na_2CO_3. How much water should be added to each 100 mL of the solution to bring down its strength to decinormal ?

Answer» <html><body><p><br/></p>Solution :The corresponding equation is: <br/> `underset(105.99 g)(Na_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)CO_(3)) + underset(98.076 g)(H_(2)SO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)) to Na_(2)SO_(4) + H_(2)O + CO_(2)` <br/> The amount of `Na_2CO_3` present in 21.2 mL solution.<br/> `=3/100 xx 21.2 = 0.63 g` <br/> `therefore 105.99 g` of `Na_(2)CO_(3)` react with `H_(2)SO_(4) = 98.076 g` <br/> `therefore 0.636 g` of `Na_(2)CO_(3)` will react with `H_(2)SO_(4)` <br/> `=98.076/105.99 xx 0.636 = 0.588 g` <br/> This much `H_(2)SO_(4)` is present in 20 mL. <br/> `therefore <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>=(NEV)/1000` <br/> `therefore 0.588 = (N xx 49.038 xx 20)/1000` <br/> (Eq. wt. of `H_(2)SO_(4) = (98.076)/2 = 49.038`) <br/> `therefore N=0.599 = 0.6` <br/> Therefore, the normality of the given `H_2SO_4` solution is 0.6 N. Suppose, v mL of water are <a href="https://interviewquestions.tuteehub.com/tag/required-1185621" style="font-weight:bold;" target="_blank" title="Click to know more about REQUIRED">REQUIRED</a> to be added to 100 mL of it to make it decinormal `N/10` <br/> `therefore 0.6 xx 100 = 1/10 xx (100 + v)` which <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> v = 500 mL<br/> Hence, 500 mL of water should be added to each 100 mL of the solution of `H_2SO_4` to bring down its strength to decinormal.</body></html>


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