The mass of $ 2.43 \mathrm{L} $ of $ \mathrm{CH}_{4} $ at $ 1.5 \mathrm{atm} $ and $ 27^{\circ} \mathrm{C} $ is
a) $ 1.6 \mathrm{g} $
b) $ 2.4 \mathrm{g} $
Gas law states that 22.4 liters of any gas at STP has gram molecular weight
CH4 is methane. It molecular weight = 12 + 4x1 = 16
So 22.4 liters of methane CH4 would weight 16 gms
The volume of Methane at 1.5 atm pressure, V2 = 2.43 liters; P2 = 1.5 atm
The volume of Methane at 1 atm =V1, pressure P1 = 1 atm
Using Boyle's law P1V1 = P2V2; V1 = P2V2/P1 = (2.43x1.5)/1 = 3.645 liters
So if 22.4 liters of methane weighs 16 gm
3.645 liters of methane would weigh 16 x 3.645/22.4 = 2.6 gm Answer
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