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34.05 mL of phosphorus vapour weigh 0.0625 g at 546^(@)C and 1.0 bar pressure. What is the molar mass of phosphorus ? |
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Answer» <P> Solution :Step 1. Calculation of VOLUME at `0^(@)C` and 1 BAR pressure`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)), i.e., (1xx34.05)/(546+273)=(1xxV_(2))/(273)"or" V_(2)=11.35 mL` 11.35 mL of vapour at `0^(@)C` and 1 bar pressure weigh `=0.0625" G"` Step 2. Calculation of mass of 22700 mL of `0^(@)C` and 1 bar pressure `:. 22700" mL"` of vapour at `0^(@)C` and 1 bar pressure will weigh `=(0.0625)/(11.35)xx22700=125" g"` `:. """Molar mass" =125" g "mol^(-1)` Alternatively, using `R=0.083" bar "DM^(3)K^(-1)mol^(-1)` PV=nRT, i.e., `n=(PV)/(RT)=(1.0" bar"xx(34.05xx10^(-3)dm^(3))/(0.083" bar " dm^(3)K^(-1)xx819 K)=5xx10^(-4)mol` `:.`Mass of 1 mole `"=(0.0625)/(5xx10^(-4))"g"=125 g` `:. ""` Molar mass`=125 g mol^(-1)` |
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