Question #90bc3

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Answer 1 · 2017-06-09T12:17:05

Given

Volume of nitrogen $V = 82.1 m L = 0.0821 L$ $V=82.1mL=0.0821L$

Pressure of the gas collected excluding aqueous tension

$P = (774.5 - 14.5) m m = \frac{760}{760} = 1 a t m$ $P=(774.5-14.5)mm=760/760=1atm$

Temperature of the gas $T = (27 + 273) = 300 K$ $T=(27+273)=300K$

Universal gas constant $R = 0.0821 L a t m K^{- 1} m o l^{- 1}$ $R=0.0821LatmK^-1mol^-1$

Let mass of the gas be $w$ $w$ g

Molar mass of the gas $M = 2 \times 14 = 28 g / m o l$ $M=2xx14=28g"/"mol$

So applying Ideal gas equation

$P V = \frac{w}{M} R T$ $PV=w/MRT$

$\Rightarrow w = \frac{P V M}{R T}$ $=>w=(PVM)/(RT)$

$\Rightarrow w = \frac{1 \times 0.0821 \times 28}{0.0821 \times 300} = \frac{28}{300} g$ $=>w=(1xx0.0821xx28)/(0.0821xx300)=28/300g$

So percentage of nitrogen in the compound

$= \frac{\frac{28}{300}}{0.14} \times 100 = 66.7 %$ $=(28/300)/0.14xx100=66.7%$