Multiple choice

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 at a 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, the masses of the individual gases in the container are respectively(assuming them to be ideal) [R = 8.314 J/mol K]

  1. 16 gm, 12 gm

  2. 4 gm, 24 gm

  3. 6 gm, 22 gm

  4. 12 gm, 16 gm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using PV = nRT, n = PV/RT = (10^5 * 0.02) / (8.314 * 300) = 0.801 moles. Let m1 be mass of Neon (20g/mol) and m2 be mass of Argon (40g/mol). m1 + m2 = 28 and m1/20 + m2/40 = 0.801. Solving gives m1 = 4g and m2 = 24g.

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