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`34.05 mL` of phosphorus vapours weighs `0.0625 g` at `546^(@)C` and `0.1` bar pressure. What is the molar mass of phossphorus ? |
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Answer» According to ideal gas equation : `PV = nRT , PV = (WRT)/(M)` or `M = (WRT)/(PV)` According to available data , Mass of phosphrous vapours `(W) = 0.0625 g` Volume of vapours `(V) = 34.05 mL = 34.05 xx 10^(-3) L` Pressure of vapours `(P) = 1.0 "bar"` Gas constant `(R) = 0.083 "bar" L K^(-1) mol^(-1)` Temperature `(T) = 546+273 = 819 K` `:. M = ((0.0625 g) xx (0.083 "bar" LK^(-1) mol^(-1)) xx (819 K))/((1.0 "bar") xx (34.05 xx 10^(-3) L)) = 125 g mol^(-1)`. |
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