At `27^@C` ina 10L flask 4.0g of an ideal gaseous mixture conatining He (molar mass 4.0 g `mol^-1`) and Ne (molar mass `20g mol^-1`) has a pressure of 1.23 atm. What is the mass% of neon? (`R = 0.082 L atm K^-1 mol^-1)`

Options

A. 84.2
B. 25.2
C. 74.2
D. 62.5

Answer

D

Explanation

Answer(4) pV = `(n_1 + n_2)RT`
pV = [W_1/M(He) + W_2/M(Ne)]RT ....(1) .
pV/RT = [W_1/M(He) + W_2/M(Ne)] .
Pressure, P = 1.23 atm Temperature, T = 27 + 273 = 300K
Volume, V = 10L
W = W_1 + W_2 =4g .
Molar mass of Helium, He = 4g/mol .
Molar mass of Neon, He = 20g/mol .
`Rigtharrow (1.23 xx 10)/0.082 xx 300 = W_1/4 + W_2/20` .
`Rightarrow 5W_1 + W_2 = 10` ....(2) Given, `W_1 + W_2 = 4` ....(3).
% mass of neon `(M_ne) = 2.5 xx 100 /4 = 62.5%`