Question

A solution is prepared by dissolving 5.64 g of glucose in 60 g of water. calculate the following:
(i) mass per cent of each of glucose and water;
(ii) molality of the solution,
(iii) mole fraction of each of glucose and water.

Answer 1

(i) Total mass of solution
$=5.64+60=65.64g$
Mass percent of glucose $=\frac{5.64}{65.64}\times 100=8.59\%$
Mass per cent of water $=(100-Masspercentofglucose)$
$=(100-8.59)=91.41\%$
(ii) No. of moles of glucose $=\frac{5.64}{180}$
Mass of water in kg $=\frac{60}{1000}$
Molality $=\frac{5.64}{180}\times \frac{1000}{60}=0.522m$
(iii) No. of moles of glucose $=\frac{5.64}{180}=0.0313$
No. of moles of water $=\frac{60}{18}=3.333$
Mole fraction of glucose $=\frac{0.0313}{3.333+0.0313}=0.0093$
Mole fraction of water $=\frac{3.333}{3.333+0.0313}=0.9907$