{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

In experiment b the block of dry ice is allowed to

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0100 atm x 101325 Pa/atm) Since all quantities are expressed in SI units, the result will also be in SI unit (kg/mol for molar mass) M = 0.00202 kg/mol This can only be the molar mass of molecular Hydrogen H2(g) 4 Question 10: (10 points) An ideal gas mixture contains equal masses of components A and B. Given that the molar mass of A is twice that of B and that the total pressure is PT = 0.900 bar, calculate PA, the partial pressure of the component A. Let us consider a mass mA of A and a mass mB of B. We know that: mA = mB . We are told that the molar mass of A is twice that of B. So, we write: MA = 2MB . The problem has to do with partial pressures and mixtures of ideal gases. Hence, we will need to use Dalton’s law, which for A is written as: PA = xAPT where xA is the mole fraction of A in the mixture and PT is the total pressure. Since we know PT, we can obtain the partial pressure of A if we know the mole fraction of A. This mole fraction can be derived from the two relationships involving masses and molar masses. Indeed, we can write the mass of A as the product of molar mass of A by the number of moles of A. Similarly for B. mA = nA x MA and mB = nB x MB Since mA = mB , we can write: nA x MA = nB x MB Then, using MA = 2MB , we conclude that nA x 2MB = nB x MB or 2nA = nB The mole fraction of A is given by: xA = nA / (nA + nB) = nA / (nA + 2nA) = 1/3 xA...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern