A mixture of 14.2 g of H2 and 36.7 g of Ar is placed in a 100.0-L container at 290 K. (b) What is the partial pressure of Ar in atmospheres?

Video transcript

Hi everyone here we have a question telling us that 15 degrees Celsius, 0.76 g of argon gas and 0. g of neon gas are mixed in a 2.3 liter container and it's asking us what are the partial pressure and mole fraction of each gas in the mixture. So first we're going to calculate our molds of argon And that's going to equal 0.76 g times one mole Divided by Aryans Mueller Mass, which is found on the periodic table and is 39.948g. And our g are going to cancel out, giving us 0. moles of Argon. And then we're gonna do the same thing for neon. So our moles of neon Equals 0.59 g times one mole over 20.1797 g. And that equals 0. moles of neon. So our temperature Equals 15°C. And we want to change that to Kelvin. So we're gonna add 273. that equals 288. Kelvin. So our mole fraction of argon is going to equal our moles of argon over the total number of moles, which will be moles of argon plus moles of neon, which equals 0. over 0. plus 0.0 to 9 to four, which equals 0.3941. And to calculate our mole fraction of neon, we can just take one minus our mole fraction of argon because the mole fractions have to equal one. So our mole fraction of neon equals one minus moles of argon. So -0.3941 equals 0.6059. So these are our mole fractions and now we can move on to our partial pressures. So we're going to use the formula PV equals N R. T. So pressure times volume equals number of moles, times our gas constant times our temperature. And we want to solve for pressure. So we're gonna divide both sides by volume. Soapy equals N. R. T. Over. We so our partial pressure of argon equals 0.01902 moles Times 0.08, Times 288.15, divided by 2.3 L, Which equals 0. atmospheres. And now we're gonna do the same for neon. So zero point 0 to 9 to four moles Time, Times 288.15 Kelvin over 2. l equals 0. Atmospheres. So those are our partial pressures. Thank you for watching. Bye