Question

A solution is prepared by mixing 1.00 gram of ethanol (C2H6O) with 100.0 gram water to give a final volume of 101 mL.

Calculate the mole fraction for the solute and solvent, and the molality of ethanol in the solution

Answer 1

Molarity=moles of solute×volume of solution

1.00g×46.08g/moles

−

1

101.0

⋅

m

L

=

0.215

⋅

m

o

l

⋅

L

−

1

Molality

=

moles of solute

kilograms of solvent

=

1.00

⋅

g

46.07

⋅

g

⋅

m

o

l

−

1

100.0

⋅

g

×

10

−

3

⋅

k

g

⋅

g

−

1

=

0.217

⋅

m

o

l

⋅

k

g

−

1

Mass percent

=

mass of solute

mass of solution

×

100

%

=

?

?

χ

the mole fraction of water

=

n

water

Total moles in solution

100

⋅

g

18.01

⋅

g

⋅

m

o

l

−

1

1.00

⋅

g

46.07

⋅

g

⋅

m

o

l

−

1

+

100

⋅

g

18.01

⋅

g

⋅

m

o

l

−

1

...and so

χ

water

≅

1.0