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The above experiment is done at NTP. The stop cock is opened for certain time and then closed. After effusion the bulb A contains 0.1 gram of D2. Find out the number of mole of H2 in the bulb B.

The above experiment is done at NTP. The stop cock is opened for certain time and then closed. After effusion the bulb A contains 0.1 gram of D2. Find out the number of mole of H2 in the bulb B.

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Grade:11

2 Answers

Pratham Sahu
33 Points
5 years ago
volume of D2 in bulb A=1.12 L
volume of h2 in bulb A=2.24L
no. of moles of D2 in bulb A=1.12/22.4=0.05 moles
no. of moles of H2 in bulb A=2.24/22.4=0.1 moles
mass of D2 in bulb A= 0.05*4= 0.2g
mass of H2 in bulb A= 0.1*2 =0.2g
After diifusion 1g of D2 remains in bulb a which means that the mass of gas diifused is (w1) 0.2 -0.1=0.1
ratio of partial pressures of D2 and H2(PD/PH) = 0.05/01=1/2
Let the weight of hydrogen diffused be w2.
therfore
r1/r2= (M2/M1)½  * PD/PH
(w1/t)/(w2/t)= (2/4)½  * 1/2
0.1/w2= 21/2/2
w2=0.14
Therfore number of moles in H2=0.14/2=0.07
 
 
Misa
13 Points
4 years ago
The above calculation is wrong. In the last fourth line  it will be 
0.1/w2   = (1/2)1/2 .1/2 = 21/2 /4
Not (1/2)1/2/2
 

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