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Chapter 18: Problem 37

What is the molarity of \(\mathrm{Na}^{+}\) in a solution of \(\mathrm{NaCl}\) whose salinity is 5.6 if the solution has a density of 1.03 \(\mathrm{g} / \mathrm{mL}\) ?

Short Answer

Expert verified
The molarity of \(\mathrm{Na}^{+}\) in the solution of NaCl with 5.6 salinity and 1.03 g/mL density is 0.959 M.

Step by step solution

01

Calculate the mass of NaCl

The salinity of the solution is given as 5.6, which means there are 5.6 grams of NaCl dissolved in 100 mL of the solution. Thus, the amount of NaCl dissolved in the solution can be calculated as: Mass of NaCl = 5.6 g
02

Convert mass of NaCl to moles

To find the number of moles of NaCl, we need to divide the mass of NaCl by its molar mass. The molar mass of NaCl is approximately 58.44 g/mol. Thus, Moles of NaCl = \(\frac{5.6 \text{ g}}{58.44 \text{ g/mol}}\) = 0.0959 mol
03

Calculate the volume of the solution

The solution has a density of 1.03 g/mL. As we know the mass of NaCl in 100 mL of the solution, we can find the mass of 100 mL of the solution: Mass of 100 mL solution = (100 mL x 1.03 g/mL) = 103 g Now, as we know the mass and density of the solution, we can calculate the volume as: Volume = \(\frac{\text{mass}}{\text{density}} = \frac{103 \text{ g}}{1.03 \text{ g/mL}}\) = 100 mL = 0.100 L
04

Calculate the molarity of Na+

The molarity of Na+ in the solution can be calculated by dividing the moles of NaCl by the volume of the solution. Since NaCl dissociates into equal amounts of Na+ and Cl- ions, the moles of Na+ will be the same as the moles of NaCl: Molarity of Na+ = \(\frac{\text{moles of NaCl}}{\text{volume of solution}} = \frac{0.0959 \text{ mol}}{0.100 \text{ L}}\) = 0.959 M So, the molarity of \(\mathrm{Na}^{+}\) in the given solution is 0.959 M.

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