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Organic Chemistry: Some Basic Principles and Techniques — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryOrganic Chemistry: Some Basic Principles and Techniques5 Marks
Question
$0.2325g$ of an organic compound was analysed for nitrogen by Duma's method $0.0317L$ of moist nitrogen was collected at $25^\circ C$ and $755.8mm$ pressure. Calculate the percentage of nitrogen. [Aq. tension of water at $25^\circ C$ is $23.8mm$ Hg.)
✓

Answer

Mass of organic compound = $0.2325g V_1 = 0.0317L, V_2 = ? P_1 = 755.8 - 23.8mm = 730mm P_2 = 760mm T_1 = 25^\circ C + 273 = 298K T_2 = 273K $$\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2}$ $\frac{730\times0.0317}{298}=\frac{760\times\text{V}_2}{273}$ $\text{V}_2=\frac{730\times0.0317\times273}{760\times298}$ $=\frac{6317.49}{226480}=0.0279\text{L}$ $22.4\text{L of N}_2 \text{ at STP weighs}=28\text{g}$ $0.0279\text{L of N}_2\text{ at STP weighs}=\frac{28}{22.4}\times0.0279\text{g}$$\%\text{ of N}=\frac{\text{Mass of nitrogen}}{\text{Mass pf organic compound}}\times100$
$\%\text{ of N}=\frac{28}{22.4}\times0.0279\times\frac{100}{\text{Mass of organic compound}}$
$=\frac{28}{22.4}\times\frac{0.0279\times100}{0.02325}$
$=\frac{78.12}{5.208}=15\%$

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