0.25g of an organic compound gave $30 c m^{3}$ of moist nitrogen at 288K and 745mm pressure. Calculate the percentage of nitrogen (aqueous tension at 288K $=$ 12.7 mm).

33.6

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23.6

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13.7

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24.5

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Solution

The correct option is C 13.7
$P_{1} =$ Atmospheric pressure $-$ Aqueous tension $= 745 - 12.7 = 732.3$ mm
$v_{1} = 30 c m^{3} = 30 m l$
$T_{1} = 288 K$
V(ml) of $N_{2}$ at STP $= \frac{P_{1} V_{1} \times 273}{760 \times T_{1}} = \frac{732.3 \times 30 \times 273}{760 \times 288}$
V(ml) of $N_{2}$ at STP $= 27.40$ ml

$% N_{2} = \frac{m o l a r m a s s o f N_{2}}{22400 m l} \times \frac{v a l u m e o f N_{2} a t S T P \times 100}{M a s s o f c o m p o u n d}$

$= \frac{28}{22400} \times \frac{27.40 \times 100}{0.25}$

$N_{2} = 13.7 %$ .

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