An organic compound was analysed by Duma-s method- 0-40 g of compound on combustion gave 50 mL of nitrogen at 27-C and 756 mm of pressure- Calculate percentage of nitrogen in the compound-

By using formula, P_1V_1T_1=P_2V_2T_2 Volume of nbsp; N_2 at STP = nbsp;P_1V_1T_1×T_2P_2 = 756300× 50 ×273760= 45.26 mL Mass of organic compound (W)=0.40 ...

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Standard VII

Physics

Measurement of Volume

An organic co...

Question

An organic compound was analysed by Duma's method, $0.40 g$ of compound on combustion gave $50 mL$ of nitrogen at $27^{\circ} C$ and $756 mm$ of pressure. Calculate percentage of nitrogen in the compound.

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Solution

By using formula,
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$Volume of N_{2} at S T P = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} = \frac{756}{300} \times 50 \times \frac{273}{760} = 45.26 mL Mass of organic compound (W) = 0.40 g % of N_{2} = \frac{28}{22400} \times \frac{V_{2}}{W} \times 100 = \frac{28}{22400} \times \frac{45.26}{0.40} \times 100 = 14.14 %$

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An organic compound was analysed by Duma's method, 0.40 g of compound on combustion gave 50 mL of nitrogen at 27∘C and 756 mm of pressure. Calculate percentage of nitrogen in the compound.

By using formula, P1V1T1=P2V2T2 Volume of N2 at STP=P1V1T1×T2P2=756300×50×273760=45.26 mLMass of organic compound (W)=0.40 g% of N2=2822400×V2W×100=2822400×45.260.40×100=14.14%

An organic compound was analysed by Duma's method, $0.40 g$ of compound on combustion gave $50 mL$ of nitrogen at $27^{\circ} C$ and $756 mm$ of pressure. Calculate percentage of nitrogen in the compound.