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JEE Main levelNumerical answerDifficulty 2 of 3ig195

The density of $H X_{2}$ (molar mass 2) at $500 K$ and $0.50 atm$ , taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , in $mg L^{- 1}$ (one decimal) is

Numerical answer type, answer in mg/L, tolerance 0.1.

Answer: 24.4 mg/L

d = \frac{PM}{RT} = \frac{0.50 \times 2}{0.0821 \times 500} = 0.02436 g L^{- 1} = 24.4 mg L^{- 1}

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