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4 grams of H2 (hydrogen) is diffused through a small hole in 5 sec at 1 atm pressure.The amount of oxygen diffused in the same interval of time at 1.5 atm and same temperatuure will be_____

5 years ago

Pallav Agarwal
33 Points

Let gas 1 be H2. Gas 2 is O2

(R1/R2)=(P1/P2)*(sqroot(m2/m1))=(n1/n2)

=> (P1/P2)*(sqroot(m2/m1))=(n1/n2)

=> (1/1.5)*(sqroot(32/2))=(2/n2)

=> (2/3)*(4)=(2/n2)

=> (n2)= 8/6   moles = 4/3 moles

=> Weight = 32*4/3 = 128/3 grams

R1 - Rate of diffusion of gas 1.

M1 - Molecular Weight of gas 1.

n1 - No. of moles of gas 1.

I am new to this forum. Please approve if you find this helpful.

5 years ago
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