Question

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.

Answer 1

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