Question

18. A 2.42 gram sample of NC15 was placed into an evacuated 2.00-L flask and allowed to partially decompose at 250.°C according to the following equation: NCls (g) NC13 (g) + Cl2 (g) The total pressure in the flask after partial decomposition was 359 torr. Calculate the mole fraction of NC13 in the flask A 0.500 B. 0.473 C. 0.540 0. 0.0540

Answer 1

Answer : B) 0.473

mas - NCIS of 2.42 g Molar mars 191.27 - mol g no. of moles mars = Molar mais 0.012 65 moles

Volume = 2.0L Temperature 250°C 273+250 = 523 K - Using the ideal gas law we get that PV = nRT P = nRT V R=0.0821 h.atm. małki P = 0.01265 x 0.0821x523 z 0.2 Tam 2.0 - NCI + Cl₂ 0.27 Huw - 0 +0 O 0.27-8 C 2

Total pressure (0.27-2) + 2+ atx 0.27 +20 We know that, I tore = 0.00132 atm . Ptotal 359 x 0.00132 atin 2 0.47 atm 0.47 = 0.27 + x => [x = 0.2 mole fraction of NCI, Puces P total 90 0.2 0.47 0.47 7 0.473
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