$\mathbf{PV}\mathbf{}\mathbf{=}\mathbf{}\mathbf{nRT}\phantom{\rule{0ex}{0ex}}\mathbf{PV}\mathbf{}\mathbf{=}\mathbf{}\left(\frac{m}{\mathrm{MM}}\right)\mathbf{RT}\phantom{\rule{0ex}{0ex}}\mathbf{MM}\mathbf{}\mathbf{=}\frac{\mathbf{mRT}}{\mathbf{PV}}$

Calculate the molar mass of a gas if 2.50 g occupies 0.880 L at 690 torr and 35 ^{o}C.

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