Answer:
The mass of nitrogen dioxide contained in a 4.32 L vessel at 48 degrees and 141600 pa is 10.54 g.
Explanation:
We know that
[tex]PV=nRT[/tex] ........... 1
Here given that
P = 141600 pa = 141.6 [tex]Kpa[/tex]
T = 48 degrees = 48 + 273 k
R = 8.314 [tex]kpa/mol k[/tex]
V = 4.32 L
Upon substituting these values in equation 1 , we get
n = 0.229 [tex]mol[/tex]
We know that molar mass of nitrogen dioxide is 46 [tex]g/mol[/tex].
So mass of 0.229 [tex]mol[/tex] become
⇒ [tex]\frac{0.229\, mol}{46 \;g/mol} = 10.54 g[/tex]
The mass of the nitrogen dioxide is 10.53 g.
The given parameters:
The number of moles of the gas is calculated by applying ideal gas law as follows;
PV = nRT
[tex]n = \frac{PV}{RT} \\\\n = \frac{141,600 \times 4.32 }{8,314 \times 321} \\\\n = 0.229 \ mole[/tex]
The mass of the nitrogen dioxide is calculated as follows;
Molar mass of nitrogen dioxide = 46 g/mole
The mass = 0.229 mole x 46 g/mole
The mass = 10.53 g
Thus, the mass of the nitrogen dioxide is 10.53 g.
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