An ideal gas weighing 9.0 g occupies a volum | vedclass

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Question

(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{w}{M}$.Substituting $n = \frac{w}{M}$ into the equation,we get $PV = \frac{w}{M}RT$

Vedclass · Accepted Answer

$27 \ g \ mol^{-1}$

Vedclass · Answer

(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{w}{M}$.Substituting $n = \frac{w}{M}$ into the equation,we get $PV = \frac{w}{M}RT$.Rearranging for molar mass $M$,we get $M = \frac{wRT}{PV}$.Given: $w = 9.0 \ g$,$V = 8.2 \ L$,$P = 1 \ atm$,$T = 300 \ K$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.Substituting the values: $M = \frac{9.0 	imes 0.0821 	imes 300}{1 	imes 8.2}$.$M = \frac{9.0 	imes 24.63}{8.2} = 9.0 	imes 3 = 27 \ g \ mol^{-1}$.

An ideal gas weighing $9.0 \ g$ occupies a volume of $8.2 \ L$ at $300 \ K$ and $1 \ atm$ pressure. The molar mass of the gas is .....

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