# How do you determine the percent by molarity and molality of 230 grams of sodium chloride dissolved in of 1025.0 g of water? Density of solution is 1.112g/mL.?

May 16, 2018

$\text{Molality} = 3.84 \cdot m o l \cdot k {g}^{-} 1$

#### Explanation:

$\text{Molality"="moles of solute"/"kilograms of solvent}$

$= \frac{\frac{230 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1}}{1025.0 \cdot g \times {10}^{-} 3 \cdot g \cdot k {g}^{-} 1} = 3.84 \cdot m o l \cdot k {g}^{-} 1$.

For $\text{molarity}$ we need to calculate the VOLUME of the resultant solution, and we have the data to do so, i.e. ${\rho}_{\text{solution}} = 1.112 \cdot g \cdot m {L}^{-} 1$

And so $\text{volume"="mass"/"density} = \frac{230 \cdot g + 1025 \cdot g}{1.112 \cdot g \cdot m {L}^{-} 1} = 1128.6 \cdot m L$

And so $\text{molarity"="moles of solute"/"volume of solution}$

$= \frac{\frac{230 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1}}{1128.6 \cdot m L \times \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1} = 3.49 \cdot m o l \cdot {L}^{-} 1$

At weaker concentrations, (approx. $< 0.1 \cdot m o l \cdot {L}^{-} 1$) molarity and molality are equivalent....