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EAMCET Engineering 2010 Solved Paper
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© examsnet.com
Question : 157
Total: 160
CH
4
diffuses two times faster than a gas
X
. The number of molecules present in
32
g
of gas
X
is (
N
is Avogadro number)
N
N
2
N
4
N
16
Validate
Solution:
From Graham's law of diffusion,
r
CH
4
r
X
=
√
M
X
M
CH
4
2
=
√
M
X
16
(given,
r
CH
4
=
2
⋅
r
X
)
M
X
=
16
×
4
=
64
Thus, the molecular mass of gas
X
is 64 .
Number of molecules of gas
X
in
32
g
gas
=
32
64
×
N
=
N
2
© examsnet.com
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