A solution of glucose in water is labelled as `10 "percent" w//w`, what would be the molality and mole fraction of each component in the solution? If the density of the solution is `1.2 g mL^(-1)`, then what shall be the molarity of the solution?

Explanation

Calculation of molality of solution
Mass of glucose in soluion = 10 g
mass of solution=100g
Mass of water in solution=(100-10)=90 g 0.09kg
`"Molality of solution (m)"=("mass of glucose/Molar mass of glucose")/("Mass of solvent in kg")`
`=(10g//(180g mol^(-1)))/((0.09 kg ))=0.617 mol//kg=0.617 m.`
Calculation of mole fractio of each component in solution.
`"No. of moles of glucose"=("Mass of glucose")/("Molar mass")=((10g))/((180g mol^(-1)))=0.055 mol`
`"No. of moles of water"=("Mass of water")/("Molar mass")=((90g))/((18g mol^(-1)))=5.0 mol`
`"Mole fraction of glucose"=(""^(n)C_(6)H_(12)O_(6))/(""^(n)C_(6)H_(12)O_(6)+""^(n)H_(2)O)=((0.055mol))/((0.055mol)+(5.0mol))=0.01`
Mole fraction of water=1-0.01=0.99
Calculation of molarity of solution=100 g
Mass of solution = 100 g
Density of solution =1.2 g `mL^(-1)`
`"Volume of solution"=("Mass of solution")/("Density")=((100g))/((1.2g mL^(-1)))=83.33 mL`
=0.08333 L.
`"Molarity of solution(M)"=("Mass of glucose //Molar mass of glucose")/("Volume of solution in litter")`
`=((10g)//(180g mol))/((0.08333 L))=0.67 mol L^(-1)=0.67 M`