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Q)

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

$(a)\;28.12\qquad(b)\;29.20\qquad(c)\;14.56\qquad(d)\;18.12$

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A)
Toolbox:
  • $Pressure = \large\frac{density\; \times RT}{molecular \;mass}$
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)
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