Question #52822

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Question #52822

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Answer 1 · 2016-10-17T00:21:44

Let the mass of CO was $w g$ $wg$

Now

$M \to Molar mass of C O = 28 g/mol$ $M->"Molar mass of "CO =28" g/mol"$

$V \to Initial Vol of CO = 0.52 L$ $V->"Initial Vol of CO"=0.52L$

$P \to Initial Pressure of CO = 1.5 a t m$ $P->"Initial Pressure of CO"=1.5atm$

$V \to Initial Temp of CO = 331 K$ $V->"Initial Temp of CO"=331K$

$R \to Universal gas constant = 0.082 L a t m K^{- 1} m o l^{- 1}$ $R->"Universal gas constant"=0.082LatmK^-1mol^-1$

By Ideal gas law

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

$\Rightarrow w = \frac{P V M}{R T} = \frac{1.5 \times 0.52 \times 28}{0.082 \times 331} = 0.805 g$ $=>w=(PVM)/(RT)=(1.5xx0.52xx28)/(0.082xx331)=0.805g$

In certain adjusted pressure and temperature when its volume becomes 2.7L ,its density will be
$D = \frac{mass}{volume} = \frac{0.805}{2.7} \frac{g}{L} \approx 0.3 \frac{g}{L}$ $D="mass"/"volume"=0.805/2.7g/L~~0.3g/L$

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Question #52822

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