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JEE Main levelNumerical answerDifficulty 2 of 3ig030

The density of a gas at $0^{\circ} C$ and $1 atm$ is $1.25 g L^{- 1}$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , its molar mass in $g mo l^{- 1}$ (one decimal) is

Numerical answer type, answer in g/mol, tolerance 0.1.

Answer: 28 g/mol

M = \frac{d RT}{P} = \frac{1.25 \times 0.0821 \times 273}{1.0} = 28.017 \approx 28.0 g mo l^{- 1}

(close to

N X_{2}

or

CO

).

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