Calculate the molality of a solution of ethanol in water in which the molefraction of ethanol is 0.040
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Explanation:
Morality (m)
Massofsolventinkg
Noofmolesofsolute
∴m=
Molecularweightofsolute(M
A
)
weightofsolute(w
A
)
×
weightofsolvent(W
B
)
1000
m=
M
A
W
A
×
W
B
1000
×
M
B
M
B
=
n
B
n
A
×
M
B
1000
n
A
= no of moles of solute m=
n
B
n
A
×
M
B
1000
×
(n
A
+n
B
)
(n
A
+n
B
)
n
B
= no of moles of solvent
∴
m=
x
B
x
A
×
M
B
1000
...(1) x
A
= mole fraction of solute
x
B
= mole fraction of solvent
Given :- x
A
=0.040 Also x
A
+x
B
=1⇒x
B
=0.960
Ethanol is solute 1 water is solvent ∴M
B
=19
Putting values of x
A
,x
B
&M
B
in (1), we get
m=
0.960
0.040
×
18
1000
⇒
m=2.31mol/kg
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