At 273 K temperature and 76 cm of Hg pres | vedclass

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(C) Given: $P = 76 \ cm \ of \ Hg = 1 \ atm$,$T = 273 \ K$,$V = 1 \ L$,$w = 1.786 \ g$.Using the ideal gas equation: $PV = nRT = (w/M)RT$.$M = (wRT) /

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Propene

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(C) Given: $P = 76 \ cm \ of \ Hg = 1 \ atm$,$T = 273 \ K$,$V = 1 \ L$,$w = 1.786 \ g$.Using the ideal gas equation: $PV = nRT = (w/M)RT$.$M = (wRT) / (PV) = (1.786 	imes 0.0821 	imes 273) / (1 	imes 1) \approx 40 \ g/mol$.However,at $STP$ $(273 \ K, 1 \ atm)$,$1 \ mole$ of an ideal gas occupies $22.4 \ L$.Thus,$M = (1.786 \ g / 1 \ L) 	imes 22.4 \ L/mol \approx 40 \ g/mol$.Checking molar masses: $2-	ext{methylpropane} (C_4H_{10}) = 58 \ g/mol$,$	ext{Propane} (C_3H_8) = 44 \ g/mol$,$	ext{Propene} (C_3H_6) = 42 \ g/mol$,$	ext{Phosgene} (COCl_2) = 12 + 16 + 71 = 99 \ g/mol$.Re-evaluating the calculation: $1.786 	imes 22.4 = 40.0064 \ g/mol$. The closest value to $40 \ g/mol$ is $42 \ g/mol$ (Propene). Given the options,$42 \ g/mol$ is the most appropriate choice.

At $273 \ K$ temperature and $76 \ cm \ of \ Hg$ pressure,the volume of $1.786 \ g$ of a gas is $1 \ L$. The gas is likely to be .........

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