Question

31.7 ml of moist N2, was obtained from 0.2033 gm of an organic compound in Duma's method at 14°C
and 758 mm pressure. If aq, tension at 14°C = 14 mm than calculate % of N
​

Answer 1

Answer:

• 0:;%%;)-5

Explanation:

@46806426852797#8::$;)&";("";:*&9$%?9%*:9&5

Answer 2

Answer:

18.15 %

Explanation:

$P_{1} = Pressure \: of \: N_{2}-Aq. \: tension \\ P_{1} = 758 - 14 \\ P_{1} = 744mm \\ P_{2} = 760mm \: T_{2} = 273 \\ V_{2} = 31.7 \\ T_{1} = 273 + 14 \\ T_{2} = 287 \\ V_{2} = \frac{P_{1} V_{1} }{T_{1}} \times \frac{273}{760} \\ V_{2} = \frac{744 \times 31.7 \times 273}{287 \times 760} \\ V_{2} = \frac{744 \times 317 \times 273}{287 \times 760 \times 10} \\ V_{2} = \frac{64386504}{2181200} \\ V_{2} = 29.51$

For percentage N

$\% \: N = \frac{28}{22400} \times \frac{V}{W} \times 100 \\ \% \: N = \frac{ \cancel28}{ \cancel22400} \times \frac{29.51}{0.2033} \times 100 \\ \% \: N = \frac{1}{800} \times \frac{2951}{0.2033} \\ \% \: N = \frac{29510000}{800 \times 2033} \\ \% \: N = \frac{29510000}{1626400} \\ \% \: N = 18.15\%$