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An ideal gas (X) present in a vessel of volume V L exerted a pressure of 16.4 atm at 200 K. What is its concentration in mol L^-1 ?
(Given $R = 0.082 Latm {mol}^{- 1} K^{- 1}$ )

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0.50

0.25

1.00

1.50

answer is C.

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Detailed Solution

$PV = nRT 16.8 \times V = 1 \times 0.082 \times 200 V = \frac{1 \times 0.082 \times 200}{16.8} = 1 L For 1 mol at V = 1 L, concentration is 1.00 mol / L$

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