Wednesday, 13 February 2013

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?


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

8 comments:

  1. well the steps are very indeed helpful and understandable....
    but over here i could find a mistake in the molality.
    0.0556/0.09=0.617 or 0.62(after rounding) not 16.7
    If you could check it .. it will be helpful .....

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  2. Sir please help
    In this question we consider 90 g of water which means that the volume of water must be 90 mL as the density of water is 1 g per mL.
    Now when we dissolve glucose in it that is 10 g of glucose then volume of the solution must increase from the initial that is 90 mL or at least remain constant . But in no case the volume should decrease. Now in this question when we solve we get the final volume as 83.3 mL, which means that the volume of the solution has finally decreased. ( note there is no temperature change given to us )How is it possible? Please tell if I am wrong somewhere. Thank you..

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  3. Sir please,
    Make it in short

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  4. Sir you have done a well explained solution sir please tell me how to do similar as you have done in other questions . I am in class 12 now and I have my boards

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