Question

How many grams of Kr are in a 7.04 L cylinder at 62.9 ∘C and 8.38 atm?

Answer 1

sol:-

• 
  R=0.0821(L x atm)/(mol. x K)
  
• atomic mass of Kr = 83.8 g/mol

data provided in the question is :-

• Volume of cylinder = 7.04 L
• Temperature = 62.9 oC = (62.9 + 273) K = 335.9 K
• pressure = 8.38 atm

​​​​​​​Formula used =

PV = nRT

where ,

P = pressure

V = Volume of cantainer

n = number of mole of gaseous substance

R = gas constant

T = temperature

calculation :-

PV = nRT

n = (PV)/(RT)

= ( 8.38 atm * 7.04 L)/{( 0.0821 L atm/mole K) * 335.9 K)

= {(8.38 * 7.04)/(0.0821 * 335.9)} mole

= (58.9952/27.57739) mole

= 2.1393 mole

n = 2.1393 mole

then,  

mass of Kr = (mole of Kr) * ( atomic mass of Kr)  

= (2.1393 mole ) * (83.8 g/mole)

= 179.27334 g

mass of Kr = 179.27334 g