A sugar syrup (containing sugar and water only) of weight 214.2 g contains 34.2 g of sugar (C12H22O11). Find the mole fraction of the sugar in the syrup.
A
0.1×10−2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
9.9×10−3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
9.9×10−4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.1×10−3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B9.9×10−3 Mass of sugar in the syrup = 34.2 g Mass of water in syrup = 214.2 - 34.2 = 180 g Molar mass of sugar (C12H22O11) = 342 g Number of moles = given massmolar mass Moles of sugar = nsugar=34.2342 = 0.1 mol Moles of water = nwater=18018 = 10 mol Mole fraction = number of molestotal number of moles xsugar=nsugarnsugar+nwater = 0.110.1=9.9×10−3