Question

A sugar syrup of weight $214.2$ g contains $34.2$ g of sugar $\left(C_{12}{H}_{22}{O}_{11}\right)$. Calculate the mole fraction of the sugar in the syrup.

(If the answer is X, then write $\frac{1000}{11}$ X)

Answer 1

${\chi }_2=\frac{n_1}{n_1+n_2}=\frac{34.2/342}{\frac{180}{18}+\frac{34.2}{342}}=0.0099$
Weight of solvent $=214.2-34.2=180g$
The molal concentration of the sugar syrup $m=\frac{W_2\times 1000}{M{w}_2\times W_1}=\frac{34.2\times 1000}{342\times 180}=0.555$
Thus, the molal concentration of the sugar syrup is $0.555$ m.