Question

calculate the mole fraction of glucose solution containing 18g of glucose in 500g of water
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Answer 1

Answer:

1

Explanation:

For glucose:

For glucose:Given mass = 18 g

For glucose:Given mass = 18 gMolecular mass =6×C+12×H+6×O =6(12)+12(1)+6(16) =72+12+96 =180 g mol^-1

For glucose:Given mass = 18 gMolecular mass =6×C+12×H+6×O =6(12)+12(1)+6(16) =72+12+96 =180 g mol^-1For water:

For glucose:Given mass = 18 gMolecular mass =6×C+12×H+6×O =6(12)+12(1)+6(16) =72+12+96 =180 g mol^Given mass = 18g Molecular mass =6×C+12×H+6×O =6(12)+12(1)+6(16) =72+12+96 =180 g mol^-1For water: Given mass = 500 gMolecular mass =2H+O =2(1)+16 =2+16 =18 g mol^-1

Now, Moles of glucose (C6H12O6)

n = m/M

n = m/Mn = 18 g/180 g mol^-1

n = m/Mn = 18 g/180 g mol^-1n = 0.1 mole

Moles of water (H2O)

n = m/M

n = m/Mn = 500 g/18 g mol^-1

n = m/Mn = 500 g/18 g mol^-1n = 27.78 mole

Total moles

= Moles of glucose + Moles of water

= Moles of glucose + Moles of water= 0.1 + 27.78

= Moles of glucose + Moles of water= 0.1 + 27.78= 27.88

Mole fraction of glucose

= Moles of glucose/Total moles

= Moles of glucose/Total moles= 0.1 mole/27.88 mole

= Moles of glucose/Total moles= 0.1 mole/27.88 mole= 0.004

Mole fraction of water

= Moles of water/Total moles

= Moles of water/Total moles= 27.78 mole/ 27.88 mole

= Moles of water/Total moles= 27.78 mole/ 27.88 mole= 1

Total mole fraction

= Mole fraction of glucose + Mole fraction of water

= Mole fraction of glucose + Mole fraction of water= 0.004 + 1

= Mole fraction of glucose + Mole fraction of water= 0.004 + 1= 1.004

= Mole fraction of glucose + Mole fraction of water= 0.004 + 1= 1.004= 1