Question

A solution of table sugar in water contains 5.25% by mass sucrose. Calculate:

1) the mass of sucrose in 225 g of solution

2) the amount of solution needed to obtain 70.0 g of sucrose

3)the mass of water contained in 450 g of the solution

4)the mole fraction of sucrose and water in 5.25% by mass sucrose.

Answer 1

(1) 100 g sucrose solution has 5.25 g of sucrose

so, 225 g of solution has mass of sucrose = $\frac{5.25}{100}\times 225 = 11.8125 g$

(2) 5.25 g of sucrose present in 100 g solution

so, 70 g of sucrose present in $\frac{100}{5.25}\times 70 = 1333.33 g$

(3) 100 g of solution has 5.25 g sucrose

450 g of solution has = 21 g sucrose

Mass of water = 429 g

(4)mass of water in 5.25 % = 94.75 g

mole of water = $\frac{94.75}{18}=5.264$

mole of sucrose = $\frac{5.25}{344}=0.015$

Mole fraction o sucrose

$\chi_1= \frac{n_1}{n_1+n_2}= \frac{0.015}{0.015+5.264}=0.0028$

Mole fraction of water

$\chi_2= 1-0.0028 =0.9971$