Calculate the weight of `N_(2)` in `6.24` litre cylinder at `27^(@)C` and `750 mm hg` `(R =0.0821` litre atm `K^(-1) mol^(-1))` .
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The correct Answer is:
7
Given `V =6.24 litre, P = (750)/(760)atm` `T = 27 + 273 = 300K, m = 28` Now `PV = (w)/(m0RT)` `:. W = (PVm)/(RT) = (750 xx 6.24 xx 28)/(760 xx 0.0821 xx 300) =7g` .
