Question

an aqueous sodium acetate, nac₂h₃o₂, solution is made by dissolving 0.415 mol nac₂h₃o₂ in 0.630 kg of water. calculate the molality of the solution.

Explanation:

Step1: Recall molality formula

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent, $m=\frac{n}{m_{solvent}}$, where $n$ is the number of moles of solute and $m_{solvent}$ is the mass of the solvent in kg.

Step2: Identify values

Here, the number of moles of solute ($n$) of sodium - acetate ($NaC_2H_3O_2$) is $n = 0.415$ mol and the mass of the solvent (water) $m_{solvent}=0.630$ kg.

Step3: Calculate molality

Substitute the values into the formula: $m=\frac{0.415\ mol}{0.630\ kg}\approx0.659\ m$.

Answer:

$0.659$