Question

calculate mole fraction of solute containing 18 grams of glucose in 100 grams of water​

Answer 1

For glucose:

Given mass = 18 g

Molecular mass =6×C+12×H+6×O =6(12)+12(1)+6(16) =72+12+96 =180 g mol^-1

For water:

Given mass = 500 g

Molecular mass =2H+O =2(1)+16 =2+16 =18 g mol^-1

Now, Moles of glucose (C6H12O6)

n = m/M

n = 18 g/180 g mol^-1

n = 0.1 mole

Moles of water (H2O)

n = m/M

n = 500 g/18 g mol^-1

n = 27.78 mole

Total moles

= Moles of glucose + Moles of water

= 0.1 + 27.78

= 27.88

Mole fraction of glucose

= Moles of glucose/Total moles

= 0.1 mole/27.88 mole

= 0.004

Mole fraction of water

= Moles of water/Total moles

= 27.78 mole/ 27.88 mole

= 1

Total mole fraction

= Mole fraction of glucose + Mole fraction of water

= 0.004 + 1

= 1.004

=1