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  • Need clarity, kindly explain! In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrog

In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is :

  • Option 1)

    16.76

  • Option 2)

    15.76

  • Option 3)

    17.36

  • Option 4)

    18.20

 

Answers (1)

best_answer

As we discussed in

Dumas method- Quantitative analyse of nitrogen -

Volume\:of\:nitrogen\:at\:STP=\frac{P_{1}V_{1}\times 2t3}{760\times T_{1}}

 


percentage\:of\:Nitrogen=\frac{28\times vol\:of\:N_{2}\:at\:STP\times 100}{22400\times m}

- wherein

 

 V_{2}\:at\:STP= \frac{P_{1}V_{1}\times 273}{T_{1}\times 760}\:ml

\%\:of\:N= \frac{28\times V_{2}\times 100}{22400\times m}

m = mass of organic compound

\%\:of\:N= \frac{58\times P_{1}v_{1}\times 2+3\times 100}{22400\times T_{1}\times 760\times m}

P1 = 725 - 25  = 700 mm

v1 = 40 ml

m = 0.25 g

T1 = 300

\%\:of\:N= \frac{28\times 700\times 273\times 100}{22400\times 300\times 760\times 0.25} = 16.76


Option 1)

16.76

This option is correct

Option 2)

15.76

This option is incorrect

Option 3)

17.36

This option is incorrect

Option 4)

18.20

This option is incorrect

Posted by

divya.saini

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