Step-1
What is given in
problem?
In this problem mass of
glucose is 10%w/w
So take mass of total
solution = 100 g
Then mass of glucose
will = 10% of 100 g = 10 grams
Mass of water will = 100
– 10 = 90 grams
And density = 1.2gmL-1
Step-2
find molality
Molality = number of
moles of solute / mass of solvent in Kg
……(1)
Here have to find number
of moles and mass of solvent in Kg both
mass of solute = mass of glucose / molar mass of
glucose …..(2)
Molar mass of glucose
(C6H12O6) = 6*C + 12*H + 6*O
= 6*12 + 12*1 + 6*16
=180 g /mol
Plug the value back in
equation (2) we get
Number of moles = 10/180
= 0.0556 moles
We have mass of solvent
(H2O) =
90 g
Mass in Kg = 90 g* 1 Kg / 1000 g = 0.09 Kg
Plug the value in
equation (1) we get
Molality
= 0.0556/0.09 = 16.7M
Step – 3
Find molar fraction
Molar fraction of
glucose = number of moles of glucose /
total number of moles
Molar mass of water
(H2O) = 2*H+ O = 2*1 + 16 = 18 g/ mol
Number of moles of
water = 90/18 = 5 moles
Plug the value in
formula we get
Mole fraction
= 0.0556/(0.0556+ 5) = 0.01
It is binary solution
because it has 2 component only so molar fraction of
Water =1 – mole fraction of glucose
= 1-0.01 =0.99
Step – 4
Find the molarity ,
molarity = number of
moles of solute / volume of solution in liter ……….(1)
Volume = mass / density
= 100 g / 1.2 g/ ml
= 83.33 ml
Mass in liter = 83.33 * 1 liter /1000 ml = 0.0833liter
Plug the value back
equation (1) we get
Molarity = 0.0556/
0.0833
= 0.67 M
