Answer:
1.6 M.
Explanation:
We want to find the molarity of a solution that contains 114.4 g NaCl in a 1.2 L solution.
Recall that molarity is given by:
[tex]\displaystyle \text{M} = \frac{\text{ mols solute}}{\text{ L soln.}}[/tex]
Convert 114.4 g NaCl to mol NaCl using its molecular weight:
[tex]\displaystyle 114.4 \text{ g NaCl} \cdot \frac{1 \text{ mol NaCl}}{58.44 \text{ g NaCl}} = 1.958\text{ mol NaCl}[/tex]
Therefore, the molarity of the solution will be:
[tex]\displaystyle \text{M} = \frac{1.958 \text{ mol NaCl}}{1.2\text{ L soln.}} = 1.6 \text{ M}[/tex]
In conclusion, the molarity of the solution is 1.6 M.
