A solution is made by mixing 72. g of chloroform (CHC13) and 38. g of heptane (C,H16).
Calculate the mole fraction of chloroform in this solution. Round your answer to 2 significant digits.
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Explanation:
Answer and Explanation:
For getting the molar fraction in this solution, we have to look up the molar mass M of each of the involved compounds:
M
C
H
3
C
O
B
r
=
122
,
95
g
/
m
o
l
M
C
H
C
l
3
=
119
,
38
g
/
m
o
l
MCH3COBr=122,95g/molMCHCl3=119,38g/mol
Now, we are going to calculate the number of moles n of each compound within the solution
n
C
H
3
C
O
B
r
=
93
g
/
122
,
95
g
/
m
o
l
=
0.76
m
o
l
e
s
n
C
H
C
l
3
=
90
g
/
119
,
38
g
/
m
o
l
=
0.75
m
o
l
e
s
nCH3COBr=93g/122,95g/mol=0.76molesnCHCl3=90g/119,38g/mol=0.75moles
Now, we have a total number of moles of 1.51 (0.75 + 0.76). Thus, the molar fraction
x
x of chloroform will be:
x
C
H
C
l
3
=
0.75
m
o
l
e
s
/
1.51
m
o
l
e
s
=
0.50
xCHCl3=0.75moles/1.51moles=0.50
The molar fraction of chloroform is thus 0.50
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