A mixture of $N e$ and $A r$ at $250$ K has a total K.E. $= 3$ KJ in a closed vessel, the total mass of $N e$ and $A r$ is $30$ g. Find mass $%$ of $N e$ in the gaseous mixture at $250$ K.

61.63

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38.37

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50

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28.3

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Solution

The correct option is A $28.3$
Let $x$ be the mass of $N e$ and $y$ be the mass of $A r$
$x + y = 30$ ------- (1)
Molar mass of $A r = 40 g$
Molar mass of $N e = 20 g$
Moles of Neon +moles of Argon = Total moles
$\frac{x}{20} + \frac{y}{40} = n$
Total $K . E . = \frac{3}{2} n R T$
Given,
$K . E . = 3 K J, T = 250 K, R = 8.314 J K^{- 1} m o l^{- 1}$
$3000 J = \frac{3}{2} (\frac{x}{20} + \frac{y}{40}) \times 8.314 \times 250$
$2 x + y = 38.48$ ------ (2)
From (1) and (2)
$y = 21.52 g$
mass % of Neon $= \frac{m a s s o f n e o n}{t o t a l m a s s} \times 100$
$= \frac{8.48}{30} \times 100 = 28.3 %$

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A mixture of Ne and Ar at 250K has a total K.E. =3 KJ in a closed vessel, the total mass of Ne and Ar is 30g. Find mass % of Ne in the gaseous mixture at 250K.

A mixture of $N e$ and $A r$ at $250$ K has a total K.E. $= 3$ KJ in a closed vessel, the total mass of $N e$ and $A r$ is $30$ g. Find mass $%$ of $N e$ in the gaseous mixture at $250$ K.