A solution is prepared by dissolving 5.64g of flucose in 60g of water. Calculate the following :
(i) mass percent of each of glucose and water
(ii) molality of the solution
(iii) mole fraction of each of glucose and water.

Explanation

(i) Total mass of solution
`=5.64 +60 =65.64g`
Mass percent of glucose `= (5.64)/(65.64)xx100 = 8.59%`
Mass percent of water = (100 - Mass percent of glucose)
`= (100-8.59)=91.41%`
(ii) No. of moles of glucose `=(5.64)/(180)`
Mass of water in kg `= (60)/(1000)`
Molality `= (5.64)/(180)xx(1000)/(60) = 0.522 m`
(iii) No. of moles of glucose `= (5.64)/(180)=0.0313`
No. of moles of water `= (60)/(18) = 3.333`
Mole fraction of glucose `= (0.0313)/(3.333+0.0313) = 0.0093`
Mole fraction of water `= (3.333)/(3.333 + 0.0313)=0.9907`.