The vapors of $1 \mathrm{~g}$ of an element occupy $2.5625 \mathrm{~L}$ exerting a pressure of $0.5 \mathrm{~atm}$ at $1000 \mathrm{~K}$. What is the molar mass (in $\mathrm{g} \mathrm{mol}^{-1}$ ) of the element? (Assume vapors follow ideal gas equation Given $\mathrm{R}=0.082 \mathrm{~L}…

Question

The vapors of $1 \mathrm{~g}$ of an element occupy $2.5625 \mathrm{~L}$ exerting a pressure of $0.5 \mathrm{~atm}$ at $1000 \mathrm{~K}$. What is the molar mass (in $\mathrm{g} \mathrm{mol}^{-1}$ ) of the element? (Assume vapors follow ideal gas equation Given $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )

Aicharya · Accepted Answer

$\mathrm{PV}=\mathrm{nRT}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT} \Rightarrow \mathrm{M}=\frac{\mathrm{mRT}}{\mathrm{PV}}$ $=\frac{(1)(0.082)(1000)}{(0.5)(2.5625)}=64 \mathrm{~g} \mathrm{~mol}^{-1}$.

The vapors of $1 \mathrm{~g}$ of an element occupy $2.5625 \mathrm{~L}$ exerting a pressure of $0.5…

Solution