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A mixture containing 2.24 L of H2 and 1.12 L of D2 at STP is taken insidea bulb connected to another bulb by a stopcock with a small opening. The second bulb is fully evacuated, the stopcock is opened for a certain time and then closed.After effusion the first bulb is found to comtain 0.1g of D2 . Find out the number of mole of H2 in second bulb.

Rohit Shukla , 7 Years ago
Grade 11
anser 1 Answers
varun

Last Activity: 7 Years ago

HELLO ROHIT,
Molecular weight of H2 = 2 g/mol
Molecular weight of D2 = 2X2 = 4 g/mol
step 1
22.4L of H2 weighs = gm molecular weight of H2
                              =  2gm => 1mol
  So 2.24L of H2 weighs = 0.2gm => 0.1 mol
 
step 2
22.4L D2 weighs = gm molecular weight of D2
                          =      4gm => 1mol
So 1.12L of D2 weighs = 4g X 1.12L/22.4L
                                   =  0.2gm
Now,
0.2gm = 0.2g/4g/mol
           =  0.05 mol
this is the simplest way to solve this question
THANK YOU

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