This problem introduces the concept of the mixing ratio, which was also introduced in problem 1 29. The mixing ratio is simply an expression of concentration as a function of the volume of the particular species per total volume in parts per million. It's a parts per million volume ratio in Problem 29. It defines the mixing ratio as the volume of the species in question at STP, divided by the volume of the total gas at STP times 10 to the six. So in the case of benzene to solve for the volume of benzene at STP, we need to take the mass of benzene and convert it to moles of benzene by dividing by the molar mass of benzene. 78.11 g per mole. Now that we have moles of benzene, we can use the conversion factor of 22.4 to convert the moles of benzene at STP to leaders of benzene at STP. Well, then, divide that by the volume of the entire gas at STP, were given the volume of the gas nodded STP as three leaders. This is the volume at a pressure and the temperature different from STP. So all we have to do is multiplied by the ratio of pressure over temperature in Kelvin and then convert it into the STP temperature and pressure my multiplied by the ratio of the temperature Um STP, which is zero Celsius or 2 73 Kelvin, divided by the standard pressure which is 7 60 tour or one atmosphere. This then will get us out of the volume at 7 48 Tour and 23 degrees Celsius and into the volume at STP. Then we simply multiply this by 10 to the sixth and we get 9.44 times 10 to the negative three parts per million volume. Now to calculate the number of moles of benzene that is president in a centimeter cubed of the entire gas mixture, we simply need thio again. Take the mass of benzene and converted into moles of benzene by dividing by the molar mass of benzene and then convert the moles of benzene into molecules have been seen by multiplying by alpha god rose number well, then divide that by the volume of the gaseous mixture, and it does not specify STP here, so it must be the three leaders that were sampled. So we'll calculate the molecules of benzene that air present per the volume specified three leaders. We can then convert the three leaders into Centimeters Cube, recognizing there are 1000 centimeters cube per leader. This then gives us 2.30 times 10 to the 11 molecules of benzene per centimeter cube of the gaseous mixture. The second part is to do the same thing for Paul you in. The only difference here is there's a different mass of tall you in and a different Moeller Mass for tall Ewing. But the set up otherwise, is exactly the same. So we will take the mass of valuing 153 10 of the negative nineties nano nano grams converted into moles by dividing by the molar Mass, and then multiply by 22.4 to get the moles into leaders at STP. Well, then divide by the volume of the gaseous mixture at STP and on the tail end. Multiply by 10 to the sixth and we get 1.37 times 10 to the negative two ppm V. Now, to calculate the molecules of Halloween that are present in a centimeter, cubed. We'll do what's similar toe what we did before. Take the mass of tall Ewing and nanograms converted into moles by dividing by the molar Mass and then multiplied by Allah. God rose number to get the molecules of valuing for the three leaders, which we then convert into Centimeters Cube by multiplying the three by 1000 and we get 3.33 times 10 to the 11 molecules of tall ewing per centimeter cubed of the gaseous mixture.