An unknown gas shows a density of 3 g per litre at 273°C and 1140 mm Hg pressure. What is the gram
molecular mass of this gas ?​

Mass of 1 litre of this gas at 273 degree Celsius and 1140 mm Hg pressure is

3 grams. So, (T1) = (273 + 273) degree Absolute = 546 degree Absolute and

(P1) = (760 + 1140) mm Hg = 1900 mm Hg and (V1) = 1 Litre

First, we have to find out how much volume does it occupy at NTP; (T2) = 273 degree Absolute and (P2) = 760 mm Hg.

So, (V2) = (1900*1 / 546)*(273 / 760) = 1.25 litre.

So, the mass of 1.25 litre volume of this gas = 3 grams

Therefore, the mass of 22.4 litre volume of this gas = (3*22.4 / 1.25) grams

= 53.76 grams.

So, the gram molecular mass of the gas is 53.76.

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c is the answer because when something is left after something evaporates that shows the chemical change.