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JEE Main Chemistry Class 11 Solutions Part 2 Questions
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© examsnet.com
Question : 58
Total: 80
Boiling point of a
2
%
aqueous solution of a non-volatile solute
A
is equal to the boiling point of
8
%
aqueous solution of a non-volatile solute
B
. The relation between molecular weights of
A
and
B
is
[27-Jul-2022-Shift-1]
M
A
=
4
M
B
M
B
=
4
M
A
M
A
=
8
M
B
M
B
=
8
M
A
Validate
Solution:
For
A
:
100
gm
solution
→
2
gm
solute
A
∴
Molality
=
2
∕
M
A
0.098
For
B
:
100
gm
solution
→
8
gm
solute
B
∴
Molality
=
8
∕
M
B
0.092
∵
(
∆
T
B
)
A
=
(
∆
T
B
)
B
∴
Molality of
A
=
Molality of
B
∴
2
0.098
M
A
=
8
0.092
M
B
2
98
×
92
8
=
M
A
M
B
1
4.261
=
M
A
M
B
∴
M
B
=
4.261
×
M
A
© examsnet.com
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