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A 13 gram sample of an unknown hydrocarbon is a gas at room temperature (25 °C) and atmospheric pressure. Its volume is 11.2 liters. Which of the following could the hydrocarbon be? Note: the ideal gas constant is 0.082 (L*atm)/(K*mol)
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Correct Answer: C. C2H2
In this question, we use the ideal gas law, PV = nRT. First of all, we calculate the number of mols in the volume of 11.2 L. n = PV/RT, or n = (1*11.2)/(0.082*298). This gives us roughly n= 0.5 mol. You can bypass this by remembering that there are 22.4 L in a gaseous mol at STP. Thus, there is 0.5 mol of a substance that we know weighs 13g. The molar mass must be 26 g/mol then. With the knowledge that C = 12g/mol and H = 1g/mol (standard knowledge), we find that the hydrocarbon should be C2H2.
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