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Calculate the weight of $N_2$ in 6.24 litre cylinder at $27^{\large\circ}C$ and 750 mm Hg. (R = 0.0821 litre atm $K^{-1}mol^{-1}$)

$(a)\;4g\qquad(b)\;7g\qquad(c)\;10g\qquad(d)\;12g$

1 Answer

Given
V = 6.24 litre
P = $\large\frac{750}{760}$ atm
T = 27+273 = 300K
m = 28
Now PV = $\large\frac{w}{m}RT$
$w = \large\frac{PVm}{RT}$
$ = \large\frac{750\times6.24\times28}{760\times0.0821\times300}$
$= 7g$
Hence answer is (b)
answered Feb 15, 2014 by sharmaaparna1
 

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