Nickel carbonyl, Ni1CO24, is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8-hr workday is 1 ppb (parts per billion) by volume, which means that there is one mole of Ni1CO24 for every 109 moles of gas. Assume 24 C and 101.3 kPa pressure. What mass of Ni1CO24 is allowable in a laboratory room that is 3.5 m * 6.0 m * 2.5 m?

Video transcript

Hi everyone for this problem. We're told that n methyl pyro Doan is a solvent used in paint and coating removal. It can be very harmful to one's health causing headache, stomach pain, nausea and vomiting. The maximum allowable concentration is 20 parts per million. We need to calculate the mass allowable in a room that has the dimensions four m 10 m and four m at 25 degrees Celsius and one A. T. M. So what we're trying to solve for here is mass. So we need to calculate the mass of this gas and we're going to need the ideal gas law in order to do that. Our ideal gas law is P. V equals N R. T. And because we need to calculate mass, what we can do with our ideal gas law is solved from moles and then from moles go to grams. Okay, so let's rearrange our ideal gas law to solve from moles. And to do that, we're going to divide both sides by R. T. When we divide both sides by R. T. We get an is equal to P. V over R. T. So, let's first solve for moles of our gas and then from molds of our gas, we can go two g of gas, which is going to give us the mass. Okay, so, our first variable is pete and equals P. V. R. T four P. We're told it is 1 80 M. And our problem. All right, so, we have our P. So we're right out here. P is equal to one ATM V. They give us the dimensions of the room and remember we can find the volume of this room by taking the cubic meter of the room and from cubic meter we can we can convert that to volume. So our volume is going to be four m times m times four meters. And that gives us 160 cubic meters. Okay, from cubic meters we can go to leaders so we can use the conversion that one cubic meter is leaders. So that gives us a volume of 160,000 leaders. So we have our volume. Our is our gas constant. This is a value. We should know when dealing with gasses at 0.08-0 six leaders times ATM. Over more times kelvin. And our temperature, they tell us that We have 25°C. And the problem. Okay, so our constant has our temperature in Kelvin. So we need to convert this temperature and C to Kelvin by adding 273.15. So we get a temperature Of 298.15 Kelvin. So we have all of our variables. We have pressure volume are constant. Anti So let's go ahead and plug into solve for moles. We get our most is equal to our pressure 1 M Times R. vol. 160,000 leaders Over our our constant 0. Leaders times a T M over more times kelvin times temperature 298.15 Calvin. Okay, so all of our units should cancel and we should be left with moles. So let's just verify that here are a T. M. S cancel. Our leaders cancel and our kelvin cancels. So we're left with moles which is perfect because that's what we're looking for. So we have, When we solve for moles, we get 0.64 malls of our gas C five H nine and oh, so now that we have our moles of gas, we need to calculate the mass of this gas and we can do that. Bye, nope. This is not the gas. Sorry, that's this is moles of gas. Okay, so now we need to convert from our moles of gas. Two moles of n. Methyl pyro don't. And we can do that by using the conversion they give us in the problem, they tell us that for One mole per 10.6 moles of gas. So let's go ahead and use that conversion here. So for every one mole of n methyl pyro Ladan, we have 10 to the six more of gas. Okay, so our units of moles of gas cancel and we're left with moles of r N. Methyl pyro dome. So we need to go from moles to grams of N. Methyl pyro don't. And we can do that using the molar mass of n. Methyl pyro don't. Okay, so we need to look up the molar mass or calculate it. So in one mall of C five H nine and O. We have a molar mass of 99.15 g. Okay, so our moles cancel and we're left with grams, which is what we're looking for here. We're looking for the mass of n methyl parallel. Don't allowable. And once we calculate that, I'm going to write it underneath here, our final answer is 0. grams of our n methyl pyro Ladan. So C five H nine and oh, okay, so that's going to be our mass. That's the end of this problem. I hope this was helpful.