Question

214.2 gram of sugar syrup contains 34.2 gram of sugar. Caluculate (i) molality of the solution and (ii) mole fraction of the sugar in the syrup-

Answer 1

(i) Mass of sugar = 34.2g
No. of moles of sugar `= (34.2)/("Mol.mass")=(34.2)/(342)=0.1`
Mass of water `=(214.2-34.2)`
`= 180g = (180)/(1000)kg`
No. of moles of water `= (180)/(18)=10`
Molality `= ("No. of moles of sugar")/("Mass of water in kg")=(0.1)/(180)xx1000`
`=0.555m`
(ii) Total number of moles `=10.0+0.1=10.1`
Mole fraction of sugar `= ("No.of moles of sugar")/("Total number of moles")`
`=(0.1)/(10.1)=0.0099`.