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A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H_(2)SO_(4). The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Answer» <html><body><p></p>Solution :Step 1. To determine the volume of `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/used-2318798" style="font-weight:bold;" target="_blank" title="Click to know more about USED">USED</a>. <br/> Volume of acid taken = <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> mL of 0.5 M `H_(2)SO_(4) = 25 mL` of 1 M `H_(2)SO_(4)` <br/> Volume of alkali used for neutralization of excess acid = 60 mL of 0.5 M NaOH = 30 mL of 1 M NaOH. <br/> Now 1 mole of `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/neutralizes-7700100" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALIZES">NEUTRALIZES</a> 2 moles of NaOH (i.e. `H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O`) <br/> `:.` 30 mL of 1 M `NaOH -= 15 mL` of 1 M `H_(2)SO_(4)` <br/> `:.` Volume of acid used by ammonia = 25 - 15 = 10 mL <br/> Step 2. To determine percentage of nitrogen.<br/> Again 1 mole of `H_(2)SO_(4)` neutralizes 2 moles of `NH_(3) :. 10 mL` of 1 M `H_(2)SO_(4) -= 20 mL` of 1 M `NH_(3)` <br/> But 1000 mL of 1 M `NH_(3)` contain nitrogen = <a href="https://interviewquestions.tuteehub.com/tag/14g-273935" style="font-weight:bold;" target="_blank" title="Click to know more about 14G">14G</a> <br/> `:.` 20 mL of 1 M `NH_(3)` will contain nitrogen `= (14)/(1000) xx 20 g` <br/> But this much amount of nitrogen is present in 0.5 g of the organic compound. <br/> `:.` Percentage of nitrogen `= (14)/(1000) xx (20)/(0.5) xx 100 = 56.0`. <br/> Alternatively, % of N can be determined by applying the following equation,<br/> `% N = (1.4 xx "Molarity of the acid" xx "Basicity of the acid" xx "Vol. of the acid used")/("Mass of substance taken")` <br/> Substituting the values of all the items in the above equation, we have, `% N = (1.4 xx 1 xx 2 xx 10)/(0.5) = 56.0`</body></html>


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