Question

A mixture of hydrogen and krypton gases is maintained in a 8.19 L flask at a pressure of 1.09 atm and a temperature of 82 °C. If the gas mixture contains 0.471 grams of hydrogen, the number of grams of krypton in the mixture is

Answer 1

Given data V=80192 P=logatm T=82°C = 82+273=355k mass of hydrogen = 0.4719 Molar mass of hydrogen = 2 g/mol mole of Hydrogen= 0.4719 -0.2355 mor 291mor Here both hydrogen and Krypton are placed together. So, there will be total mol ie sum of more of hydrogen and Krypton. using ideal gas equation: PV=RT MT= PV = 1.0gath x 8.194 RT 0.082057 Kaps mortetx355x nr = 0.30 64 mor Nr = Hot nir 0.3064 = 0.2355 +nke Nkr = 0.0709 mor = morof. Krypton Molar mass of Krypton = 83.798 g/mol mass of Krypton = 0.0709 moix 83.798 g/mol = 5.9412 g Hence, no. of grams of Krypton in the mixture is 15.9412g.