Molality of an aq solution whose R.L.V.P is 0.1
Answers
Answered by
0
Answer:
Explanation:
According to Raoult's law,
Relative lowering of vapour pressure = mole fraction of solute.
(X
B
)=0.0125
∵X
B
=
1000+m⋅M
A
m⋅M
A
Where m=molality,
∴0.0125=
1000+m×18
m×18
12.5+0.225 m=18 m
17.775 m=12.5
m=
17.775
12.5
=0.70.
Hence, option A is correct.
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