Question

60 g of sucrose and 90 g of glucose are dissolved in 1000ml of solution (aqueous). The specific gravity of the resulting solution is 1.1 g/ml. The percentage of moles per sucrose present in the solution will be:

(Assume volume of solution does not change by addition of solutes)

• A. 0.328
• B. 0.258
• C. 0.469
• D. 0.152

Answer 1

The correct option is A 0.328

Mass of the solution= $1.1\times {10}^3=1100g$

$\therefore$ Mass of $H_{2}O=1100-(60+90)=950g$

Moles of sucrose= $\frac{60}{342}=0.175$, moles of $H_{2}O=\frac{950}{18}=52.78$

Moles of glucose= $\frac{90}{180}=0.5$

$\therefore$ Mole % of sucrose= $\frac{0.175}{0.175+52.78+0.5}\times 100=0.328$