Chlorine occurs as a mixture of two isotopes, 35Cl and 37Cl. What is the ratio of the diffusion rates of the three species 135Cl22, 35Cl37Cl, and 137Cl22?

Video transcript

Welcome back everyone. We're told that broom in 79 81 are isotopes of roaming. How do the diffusion rates of our di atomic molecules of bromine 79 81 our compound gas of bromine 79 81 isotopes compare. So we're going to first recall what fusion is, recall that a fusion describes our gas particles traveling through a small space. And it can be calculated by the rate of a fusion of our gas, which is inversely proportional to the square root of its molar mass. So making note of our molar mass. First beginning with our bromine, 79 di atomic molecules, we would take that mass number of 79 multiply it by two. Which will give us a mass number. Or sorry, a molar mass of 158 g per mole. Next we have our molar mass of our bruning 81 di atomic molecules where we take that 81 mass number multiply it by two And we're going to have a molar mass of 162 g per mole. Now moving on to a molar mass of our compound gas being mean 79 and 81, we have a molar mass of the two Mass numbers added together 79 plus 81 which will give us a total of 160 g per mole. So calculating our first diffusion rate, we'll begin with our bromine gas to our highest mass of our gas compound, which is associated with our di atomic bromine gas in 81. So what we'll say is that our rate of effusion of our browning 79 di atomic molecule divided by the rate of effusion of our browning 81 di atomic molecule is equal to the square root of the molar mass of our bro. Mean 79 di atomic molecules Divided by the molar mass of our browning, 81 di atomic molecules and plugging in those values. So let's make some room here, plugging in these values. We can set this equal to being the square root of Our higher value for the mass in the numerator. So we would have 162 divided by 158 for our molar mass of our bromine, 79 di atomic molecule in the denominator because it's smaller. And so this is going to simplify to a value of 1.012. So now following the same steps. But using our compound gas here, Comparing it to our highest gaseous molar mass, being our bromine 81. We're going to write a relationship where we would say that our rate of our rate of effusion of our compound gas bromine 79 Times, bringing 81 Divided by our rate of effusion of our roaming 81 di atomic molecule is equal to the square root of our molar mass of our 81 di atomic molecule divided by the molar mass of our Compounds gas being are roaming 79 and 81. And so we want that in the denominator again because it's the lower value. So we have our higher value being our molar mass of our bromine, 81 di atomic molecule of 162 Divided by R. Mueller mass of our compound gas being 160. And this is going to simplify to 1.006. Now just to be clear, we wrote our relationship here in the opposite way. So what we had here was our molar mass of our roaming 81 di atomic molecule in the numerator. Because it's the larger value where our molar mass of our booming 79 di atomic molecule was plugged in the denominator since it was a smaller value. So now that we have these effusion rates, we now can rank our rate of diffusion of these gasses where we would say that and we'll make some more room here. So we're going to say therefore the highest rate of diffusion associated with our roman 79 di atomic molecules Which had a value of about 1.013 when we rounded up is going to be the greatest rate of diffusion because it was the gas molecule with the lowest molar mass. We can see Where we have. Then up next in our ranking our compound gas bromine 79 and bromine 81 which have a rate of diffusion. And let's make this a little bit smaller so we have enough room, Which we calculated to be 1.06 because it has the second lowest molar mass so it's able to travel faster or diffuse rather faster. And this is then greater than the rate of diffusion of our broom in 81 which was a di atomic molecules that had the highest smaller mass, which will say has a rate of diffusion of one point oh. And this ranking would be our final answer describing the relative rates of diffusion of each of our gasses. I hope everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.