Calculate the mole fraction, molality and molarity of 10% of glucose(C6H12O6)
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10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 - 10) g = 90 g of water.
Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1
Then, number of moles of glucose = 10 / 180 mol
= 0.056 mol
∴ Molality of solution = 0.056 mol / 0.09kg = 0.62 m
Number of moles of water = 90g / 18g mol-1 = 5 mol
Mole fraction of glucose (xg) = 0.056 / ( 0.056+5) = 0.011
And, mole fraction of water xw = 1 - xg
= 1 - 0.011 = 0.989
If the density of the solution is 1.2 g mL - 1, then the volume of the 100 g solution can be given as:
= 100g / 1.2g mL-1
= 83.33 mL
=83.33 x 10-3 L
∴ Molarity of the solution = 0.056 mol / 83.33 x 10-3 L
= 0.67 M
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