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A closed container of volume $0.02 m^{3}$ contains a mixture of neon and argon gases, at a temperature of $27^{0} C$ and pressure of $1 \times 10^{+ 5} N m^{- 2} .$ The total mass of the mixture is $28 g$ . If the gram molecular weights of neon and argon are 20 and 40 respectively. Find the masses of the individual gases in the container, assuming them to be ideal. (Universal gas constant R=8.314 J/mol.K)

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$m_{1} = 4 g, m_{2} = 24 g$

$m_{1} = 8 g, m_{2} = 24 g$

$m_{1} = 4 g, m_{2} = 12 g$

$m_{1} = 6 g, m_{2} = 12 g$

answer is A.

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Detailed Solution

$m_{1} + m_{2} = 28; M_{1} = 20 g, M_{2} = 40 g; V_{1} + V_{2} = 0.02;$

$m_{1} = M_{1} V_{1}, m_{2} = M_{2} V_{2}$

$P_{1} V_{1} = \frac{m_{1}}{20} {RT}_{1} & P_{2} V_{2} = \frac{m_{2}}{40} {RT}_{2}$

On solving the above equations,

$m_{1} = 4 g, m_{2} = 24 g$

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