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Question
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?
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Solution 1
Let mass of solution = 100 g
∴ Mass of glucose = 10 g
Mass of water = 100 – 10 = 90 g = 0.09 kg
No. of moles in 10 g glucose = `10/180`
= 0.0555 mol
No. of moles in 90 g H2O = `90/18`
= 5 moles
Volume of solution = `(100 g)/(1.2 g mL^(-1))`
= 83.33 mL
= 0.0833 L
Molality = `"Number of moles of solute"/"Mass of solvent in kg"`
= `(0.0555 mol)/(0.09 kg)`
= 0.617 m
x (Glucose) = `"Number of moles of solute"/"Number of moles of solution"`
= `0.0555/5.0555`
= 0.01
∴ x (H2O) = 1 − 0.01 = 0.99
Molarity = `"Number of moles of solute"/"Volume of solution in L"`
= `0.0555/0.0833`
= 0.67 M
Solution 2
i. 10% w/w means that 100 g of the solution contains 10 g of glucose. Thus, the mass of water present is 100 − 10 = 90 g.
Moles of glucose present = `10/180`
= 0.055
Mass of solvent (water) = 90 g
= 0.090 kg
∴ Molality (m) = `0.055/0.090`
= 0.617 m
ii. Moles of water present = `90/18`
= 5
∴ Mole traction of glucose = `0.055/(0.055 xx 5)`
= 0.011
Mole fraction of water = `5/(0.055 xx 5)`
= 0.989
iii. Volume of 100 g solution = `"Mass"/"Density"`
= `100/1.2`
= 83.33 mL
Hence, molarity (M) of the solution
M = `(w xx 1000)/(M' xx v)`
= `(10 xx 1000)/(180 xx 83.33)`
= 0.67 M
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