A solution is made by dissolving 14.57g of sodium bromide in 415g of water. What is the molality of the solution?

1 Answer
Jun 27, 2016

$\text{Molality"="Moles of solute"/"Kilograms of solvent}$

Explanation:

For your problem,

$\text{Molality} = \frac{14.57 \cdot g}{102.89 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{0.415 \cdot k g}$ $\cong$ $0.342 \cdot m o l \cdot k {g}^{-} 1$.

Would the $\text{molarity}$ of the solution be different?