A thin glass tube 1 m long is filled with Ar gas at 101.3 kPa, and the ends are stoppered with cotton plugs as shown below. HCl gas is introduced at one end of the tube, and simultaneously NH3 gas is introduced at the other end. When the two gases diffuse through the cotton plugs down the tube and meet, a white ring appears due to the formation of NH4Cl1s2. At which location—a, b, or c—do you expect the ring to form?

Video transcript

Hi everyone here we have a question telling us that ammonia and bromide can be prepared by the direct action of hydrogen bromide on ammonia, ammonia plus hydrogen bromide forms ammonium bromide. Now consider the following experiment, a five ft tube filled with crypto At 1.0 atmosphere, with the ends covered with a rubber stopper, ammonia and hydrogen bromide were then simultaneously introduced at opposite ends of the tube. A ring of solid forms where the gasses meet due to the mentioned reaction approximately where in the tube would you expect the ring to form? So the molar mass of ammonia Equals 14 g per mole plus one gram per mole, times three Equals 17 g per mole. Our molar mass of hydrogen bromide Equals one g per mole plus 79 .9 g per mole Equals 80.9 grams. Permal graham's Law of effusion states that the rate of effusion of the gasses inverse nelly proportional to the square root of the molar mass of its particle. If both gasses move at the same rate, they would meet in the middle. Otherwise they would meet closer to the side of the slower gas, hydrogen bromide is heavier. Therefore would be slower, meaning that the gasses would meet closer to the side it was introduced, which is side three. So our answer here is three. Thank you for watching. Bye