A sugar syrup of mass 214.2g contains 34.2 of sugar .calculate the concentration of sugar in the syrup
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Answered by
68
Mass of syrup = 214.2 g
Mass of sugar =34.2 g
Molar mass of sugar = 342 g/ mol
No. of moles of sugar (nA) =34.2/342 = 0.1 mole
Mass of water in sugar syrup = 214.2-34.2 = 180 g
No. of moles of water(nB) = 180/18 =10 moles
Mole fraction of sugar (xA) = nA/ (nA + nB) =0.1/ (0.1 +10) =0.1 /10.1 =0.0099
Mass of sugar =34.2 g
Molar mass of sugar = 342 g/ mol
No. of moles of sugar (nA) =34.2/342 = 0.1 mole
Mass of water in sugar syrup = 214.2-34.2 = 180 g
No. of moles of water(nB) = 180/18 =10 moles
Mole fraction of sugar (xA) = nA/ (nA + nB) =0.1/ (0.1 +10) =0.1 /10.1 =0.0099
Answered by
65
Concentration of sugar = (Mass of sugar / Mass of syrup) × 100
= (34.2 / 214.2) × 100
= 15.96%
Concentration of sugar in given syrup is 15.96% by mass.
= (34.2 / 214.2) × 100
= 15.96%
Concentration of sugar in given syrup is 15.96% by mass.
Similar questions
Weight of sugar in syrup=34.2g
∴∴ Weight of water in sugar syrup=(214.2-34.2)g
⇒180.0g⇒180.0g
Molecular weight of sugar=(C12H22O11)(C12H22O11)=342
∴∴ Molal concentration =34.2×1000342×18034.2×1000342×180=0.56=0.56
(ii) Mole fraction of sugar
34.2/342(180/18)+(34.2/342)34.2/342(180/18)+(34.2/342)
(Mol.wt of water=18)
⇒0.110+0.1=0.110.1⇒0.110+0.1=0.110.1=0.0099=0.0099
Hence 0.56,0.0099 is the correct .