Question

A chemist prepares a solution of sodium chloride (NaCl) by measuring out 82.2g of sodium chloride into a 250. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's sodium chloride solution. Round your answer to 3 significant digits.

Answer 1

Answer:

About 5.64 mol/L.

Explanation:

Recall that the concentration of a soluion, or its molarity, is given by:

$\displaystyle \text{M} = \frac{\text{ mols solute}}{\text{ L soln.}}$

In this case, our solute is 82.2 g of NaCl and our solution is in a 250. mL volumetric flask.

Convert 82.2 g of NaCl to moles of NaCl. The molecular weight of NaCl is:

$\displaystyle \begin{aligned} \mathcal{M}_{\text{NaCl}} & = (22.99 + 35.45) \text{ g/mol} \\ \\ & = 58.44 \text{ g/mol}\end{aligned}$

Hence:

$\displaystyle 82.2\text{ g NaCl} \cdot \frac{ 1 \text{ mol NaCl}}{58.44 \text{ g NaCl}} = 1.41 \text{ mol NaCl}$

Convert the volume to liters:

$\displaystyle 250. \text{ mL} \cdot \frac{1 \text{ L}}{1000 \text{ mL}} = 0.250 \text{ L}$

Hence, the concentration of the solution is:

$\displaystyle \begin{aligned} \text{M} & = \frac{(1.41 \text{ mol NaCl})}{(0.250 \text{ L soln.})} \\ \\ & =5.64 \text{ mol/L} \end{aligned}$

In conclusion, the concentration of the solution in mol/L is about 5.64 mol/L.