For a gaseous mixtue of 2.41g of helium and 2.79g of neon in an evacuated 1.04 dm^(3) container at 298 K Calculate the partial pressure of each gas and hence find the total pressure of the mixture.

For a gaseous mixtue of 2.41g of helium and 2.79g of neon in an evacua

1.	For a gaseous mixtue of 2.41g of helium and 2.79g of neon in an evacuated 1.04 dm^(3) container at 298 K Calculate the partial pressure of each gas and hence find the total pressure of the mixture.
Answer» Answer :Mass of He = 2 . 41 g No. of moles of He = `("Mass")/("Molar Mass") = (2.41)/(4) = 0 . 6 0 25 `moles Mass Of Ne= 2 . 79 g No. of moles of Ne ` = ("Mass")/("Molar Mass") = (2.79)/(20)` = 0 . 1395 moles Volume of the Total no. of moles of the mixtureConysinrt V= 1 . 0 4 `dm^(3)` TEMPERTURE T= 298 K Pressure P` = (1)/(V) RT` ACCORDING to ideal gas equation PV = nRT ` P = ( 0 7 4 2 0 XX 0 . 0 821 xx 298 )/( 1 . 0 4) = 17 . 45` atm Partial pressure P = molefraction ` xx`Total pressure ` = (nA)/(nA + nB) xx P ` Partial Pressure of Helium ` = P_(He) = (0 . 6025)/(0.7420) xx 17 . 45 ` According to Dalton's law of partial pressure = 3 . 280 atm `P = P_(1) + P_(2) + P_(3). . . . ` ` P_("total") = P_(He) + P_(Ne) = 14. 169 + 3 . 280 = 17 . 449 ` atm