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KCET 2014 Chemistry Solved Paper
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© examsnet.com
Question : 57
Total: 60
1.78 g of an optically active L-amino acid (A) is treated with
N
a
N
O
2
/
H
C
l
at 0 °C. 448
c
m
3
of nitrogen was at STP is evolved. A sample of protein has 0.25% of this amino acid by mass. The molar mass of the protein is
34,500 g
m
o
l
−
1
35,600 g
m
o
l
−
1
36,500 g
m
o
l
−
1
35,400 g
m
o
l
−
1
Validate
Solution:
Mass of L-amino acid = 178 g
R
−
C
|
NH
2
H
−
C
O
O
H
N
a
2
/
HCl
–
–
–
–
–
–
–
–
–
>
0
°
C
N
2
At STP,
1
mol
=
22400
c
m
3
1
cm
3
=
1
22400
mol
So,
448
c
m
3
N
2
is evolved when
448
22400
mol of amino acid reacted
Now, for L-amino acid,
Molar mass of amino acid
=
22400
×
1.78
×
10
2
448
=
178
2
=
89
g
m
o
l
−
1
Sample of protein contains 0.25% amino acid, so 100 g of protein contain 0.25 g of amino acid. Therefore,
Molecular mass of protein
=
100
×
89
0.25
=
35600
g
mol
−
1
© examsnet.com
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