What is the density (in grams per liter) of a gas at STP if 1.97 L of the gas in a bulb at 23.8 °C, and 761.9 mm Hg weighs 0.371 g?

Answer 1

The easy way to do this is not the easy way to explain it. I will do it both ways. The easy way to explain first. The density under the conditions listed is g/L = 0.371/1.97L = 0.188 g/L. The general gas equation can be modified for density as PM = dRT where M is molar mass and d is density. You know d,R, T, and P, solve for molar mass.
M = dRT/P = 0.188*0.08206*297*760/761.9. Knowing M, plug that back into
PM = dRT and substitute 1 atm for P and 273.2 for T and solve for new density.

Now for the tougher way to explain but the easy to do solve it (stay with me). You know the density is 0.188 at 761.9 mm pressure and 297 K. So you want this at STP which means you must make two corrections; i.e., one for P and one for T.
0.188 g/L x (p corr'n) x (T corr'n) = ?
p correction is from 761.9 mm to 760 mm. That's a decrease in P, you know volume goes up if P goes down, and volume going up (in the denominator of g/v) means density goes down so the factor must be smaller than 1. So the P correction is 760/761.9.

The T goes from 297 to 273.2 which is a decrease in T, decreasing T means decreasing volume, and a smaller number in the denominator means a larger density so that factor must be > 1. T factor is 297/273.2. Therefore, new density is
0.188 x (760/761.9) x (297/273.2) = ?
You should get the same answer either way. I did.