Saved Bookmarks
| 1. |
Exercise 4.A certain mixture of helium and argon weighing 5 gocepressure. What is the composition of the mixture in mapies a volume of 10 L at 25°cmixture in mass percentage? |
|
Answer» Answer: Ideal Gas LAW : PV = nRT No. of moles (n) = N1 + n2 where n1 = No. of moles of Helium n2 = No. of moles of Argon 1 atm x 10 L = (n1 + n2) x 0.0821 x (273+25)
n1 + n2 = 0.4087 ..........................................(1) No. we know No. of moles x Molar mass = given mass Therefore (Molar mass x No. of moles of He ) +(Molar mass x No. of moles of argon ) = Mass of sample 4n1 + 39.9 n2 = 5 .........................................(2) [SOLVING equation 1 and 2 simultaneously for the value of n1 and n2 as follows :] 4 x (n1 + n2 = 0.4087 ) .............(3) 1 x (4n1 + 39.9 n2 = 5 )................(4) [subtracting eq 3 from 4 we get] 35.9 n2 = 3.3652 n2 = 0.0937 moles of Argon Mass of argon present = 0.0937 x 39.90 = 3.7386 g Mass of helium present = 5g - 3.73863g = 1.2613 g Mass % of He = (1.2613/5) x100 = 25.2274 % Helium Mass % of Ar = (3.7386/5) x 100 = 74.772 % of Argon |
|