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 will be the molarity of the solution ?

Explanation

0 % w / w solution of glucose in water means that 10 g 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 `(C_(6)H_(12)O_(6))`
`= 6xx12+12xx1+6xx16`
`= 180 g mol^(-1)`
Then, number of moles of glucose
`= (10)/(180)mol`
= 0.056 mol
Molarity of solution `= (0.056 mol)/(0.09 kg)=0.62 m`
Number of moles of water `= (90 g)/(18 g mol^(-1))=5 mol`
Mole fraction of glucose `(x_(g))=(0.056)/(0.056+5)=0.011`
And, mole fraction of water `X_(w)=1-x_(g)`
`=1-0.011`
= 0.989
If the ensity of the solution is 1.2 kg `mL^(-1)`, then the volume of the 100 g solution can be given as,
`= (100 g)/(1.2 g mol^(-1))`
= 83.33 mL
`= 83.33xx10^(-3)L`
`therefore` Molarity of the solution `= (0.056 mol)/(83.33xx10^(-3)L)`
= 0.67 M.