## General Chemistry: Principles and Modern Applications (10th Edition)

A saturated aqueous solution of $NaBr$ at 20$^{\circ}$C contains: $53.7g$ $NaBr$ $/ 100g$ $Solution$
1. The mass of the solution, in this is case, is the sum of the NaBr and the water mass. $mass(Solution) = 116g + 100g =216g$ 2. Now we know that there is 116g of NaBr for 216g of solution, and we just need to calculate this for 100 g of solution using a proportion. $\frac{116g}{216g} = \frac{x}{100g}$ $0.537 = \frac{x}{100g}$ $x = 0.537 \times 100g = 53.7g$