In the quantitative determination of nitrogen- N2 gas liberated from 0-42 g of a sample of organic compound was collected over water- If the volume of N2 gas collected was 10011 ml at total pressure 860 mm Hg at 250 K- - by mass of nitrogen in the organic compound is - -Aqueous tension at 250 K is 24 mm Hg and R-0-08 L atm-mol K-

The correct option is A 103% The volume of dry nitrogen at STP is V_o=P_1V_1T_1×T_oP_o=(860 24) ×1011250×273760=10.92 ml.Percentage by mass of nitro ...

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Standard IX

Physics

Atmospheric Pressure

In the quanti...

Question

In the quantitative determination of nitrogen, $N_{2}$ gas liberated from $0.42$ g of a sample of organic compound was collected over water. If the volume of $N_{2}$ gas collected was $\frac{100}{11}$ ml at total pressure $860$ mm Hg at $250$ K, $%$ by mass of nitrogen in the organic compound is :
[Aqueous tension at $250$ K is $24$ mm Hg and $R = 0.08$ L atm/mol K]

\frac{10}{3} %

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\frac{5}{3} %

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\frac{20}{3} %

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\frac{100}{3} %

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Solution

The correct option is A $\frac{10}{3} %$
The volume of dry nitrogen at STP is $V_{o} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{o}}{P_{o}} = \frac{(860 - 24) \times \frac{10}{11}}{250} \times \frac{273}{760} = 10.92$ ml.
Percentage by mass of nitrogen $= \frac{28 \times V_{o} \times 100}{22400 \times m} = \frac{28 \times 10.92 \times 100}{22400 \times 0.42} ≃ \frac{10}{3} %$ .

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The correct option is A 103%The volume of dry nitrogen at STP is Vo=P1V1T1×ToPo=(860−24)×1011250×273760=10.92 ml.Percentage by mass of nitrogen =28×Vo×10022400×m=28×10.92×10022400×0.42≃103%.