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  • In Duma's method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the

In Duma's method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be :

(Aqueous tension at 300 K = 15 mm)

Option: 1

14.45


Option: 2

15.45


Option: 3

16.45


Option: 4

17.45


Answers (1)

best_answer

From 715mm pressure, subtract aqueous tension 15mm to obtain pressure of nitrogen.

P = 715 - 15 = 700 mm Hg

Volume of nitrogen at STP   =    

                                                      \frac{V \times P \times 273}{(T) \times 760}

Volume of nitrogen at STP =  

                                                    \frac{55 \times 700 \times 273}{(300) \times 760}=46 \mathrm{mL}

Percent of nitrogen =  

                                         \frac{\text {vol of } N_{2} \text { at } S T P}{\text {wt of organic compound}} \times \frac{28}{22400} \times 100

Percent of nitrogen =  

                                          \frac{46}{0.35} \times \frac{28}{22400} \times 100=16.45 \%                                     

The percentage composition of nitrogen in the compound would be 16.45 %.

Posted by

Riya

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