At 300 , K temperature and 1 , atm pressure,t | vedclass

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(A) Using the ideal gas equation $PV = \frac{m}{M}RT$,where $M$ is the molar mass.Given: $P = 1 \, atm$,$V = 4.1 \, L$,$m = 7.0 \, g$,$T = 300 \, K$,a

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(A) Using the ideal gas equation $PV = \frac{m}{M}RT$,where $M$ is the molar mass.Given: $P = 1 \, atm$,$V = 4.1 \, L$,$m = 7.0 \, g$,$T = 300 \, K$,and $R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.Rearranging for $M$: $M = \frac{mRT}{PV}$.Substituting the values: $M = \frac{7.0 	imes 0.0821 	imes 300}{1 	imes 4.1}$.$M = \frac{172.41}{4.1} \approx 42.05 \, g/mol$.Thus,the molar mass is approximately $42 \, g/mol$.

At $300 \, K$ temperature and $1 \, atm$ pressure,the volume of $7.0 \, g$ of a gas is $4.1 \, L$. What is the molar mass of the gas?

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