0.216g of an organic compound gives 28.8 $c m^{3}$ of moist nitrogen measured at 288K and 732.7mm pressure. The percentage of nitrogen in the substance is approximately: (Aq. tension at 288K $=$ 12.7mm)

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Solution

The correct option is C 15
$P_{1} = 732.7 m m$
$V_{1} = 28.8 c m^{3} = 28.8 m l$
$T_{1} = 288 K$
at STP
$P_{2} = 760 m m$
$V_{2} = ?$
$T_{2} = 273 K$
$\frac{p_{1} v_{1}}{T_{1}} = \frac{p_{2} v_{2}}{T_{2}}$
aq. Tension $= 12.7 m m$
so $P_{1} = 732.7 - 12.7 = 720 m m$
$V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}} = \frac{720 \times 28.8 \times \times 273}{760 \times 288}$
$∴ V_{2} = 25.86 m l .$
% $N_{2} = \frac{28}{22400} \times \frac{v_{2}}{m} \times 100$
$= \frac{200}{22400} \times \frac{25.86}{0.216} \times 100$
$N_{2} =$ 14.965% (15%)

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0.216g of an organic compound gives 28.8cm3 of moist nitrogen measured at 288K and 732.7mm pressure. The percentage of nitrogen in the substance is approximately: (Aq. tension at 288K = 12.7mm)

The correct option is C 15P1=732.7mmV1=28.8cm3=28.8mlT1=288Kat STPP2=760mmV2=?T2=273Kp1v1T1=p2v2T2aq. Tension=12.7mmso P1=732.7−12.7=720mmV2=P1V1T2P2T1=720×28.8××273760×288∴V2=25.86ml.%N2=2822400×v2m×100=20022400×25.860.216×100N2=14.965% (15%)

0.216g of an organic compound gives 28.8 $c m^{3}$ of moist nitrogen measured at 288K and 732.7mm pressure. The percentage of nitrogen in the substance is approximately: (Aq. tension at 288K $=$ 12.7mm)