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Densities of air in chandigarh was found to be $1.156kgm^{-3}$. The temperature and pressure were recorded as $20^{\large\circ}C$ and $100.142\; KPa$. The average molecular mass of the air at that place in $g \;mol^{-1}$ is


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  • $Pressure = \large\frac{density\; \times RT}{molecular \;mass}$
Since $P = \large\frac{dRT}{m}$
$m = \large\frac{dRT}{P}$
$m = \large\frac{1.156\times8.314\times293}{100.142}$
$=28.12 g\;mol^{-1}$
Hence answer is (a)
answered Mar 2, 2014 by sharmaaparna1
edited Mar 18, 2014 by mosymeow_1

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