# Answer to Question #35335 in Physical Chemistry for Delaney

Question #35335

A 800mL container holds 2.366g of an unknown gas at 78.0 degrees Celsius and 103kPa. Identify the gas.

Expert's answer

Based on the ideal gas law the molecular mass of the gas was obtained in the following way:

PV = nRT; T = 351 K;

Then,

n = m/M = (PV)/(RT);

From here we determine

M = (mRT)/(PV) = (2.366*10

Therefore, the closest match is Krypton that has a molecular weight of 83.798 g/mol. Its boiling point is 119.93 K that is consistent with our

data as it's supposed to be a gas at 351K.

PV = nRT; T = 351 K;

Then,

n = m/M = (PV)/(RT);

From here we determine

M = (mRT)/(PV) = (2.366*10

^{-3}*8.31*351)/(103*10^{3}*0.8*10^{-3}) = 0.0837kg/mol = 83.7 g/molTherefore, the closest match is Krypton that has a molecular weight of 83.798 g/mol. Its boiling point is 119.93 K that is consistent with our

data as it's supposed to be a gas at 351K.

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