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If density of vapours of a substance of molar mass 18 gm/mole at 1 atm pressure and 500 K is 0.36 kg m − 3 , then value of Z for the vapours is (Take R = 0.082 L atm mole − 1 K − 1 )

Options

  1. A41 50
  2. B50 41
  3. C1.8
  4. D0.9

Correct answer

B. 50 41

Step-by-step solution

V r e a l = m o l a r m a s s d e n s i t y = 18 0.36 V i d e a l = n R T P = 1 × 0.082 × 500 1 So, Z = V r e a l V i d e a l = 50 0.082 × 500 = 50 41

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