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JEE Advanced levelSingle correct optionDifficulty 3 of 3ev047

A 250 L sample of air measured at STP contains 1.10 mg of $SO X_{2}$ (assume ideal gas, molar volume 22.4 L/mol at STP, $S = 32$ , $O = 16$ ). The concentration of $SO X_{2}$ in ppm by volume is

A4.40 ppm
B1.54 ppm
C15.4 ppm
D0.0154 ppm

Answer: B

Moles of

SO X_{2}

=

1.10 \times 1 0^{- 3} /64 = 1.72 \times 1 0^{- 5}

mol, occupying

1.72 \times 1 0^{- 5} \times 22.4 = 3.85 \times 1 0^{- 4}

L. ppm by volume =

3.85 \times 1 0^{- 4} /250 \times 1 0^{6} = 1.54

. The value 4.40 is just mass per volume in

mg m^{- 3}

.

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