# Question #87882

Sep 3, 2017

The concentration is 0.614 mol/L.

#### Explanation:

Assume you have 1 L of seawater.

Step 1. Calculate the mass of the seawater

$\text{Mass of seawater" = 1000 color(red)(cancel(color(black)("mL seawater"))) × "1.025 g seawater"/(1 color(red)(cancel(color(black)("mL seawater")))) = "1025 g seawater}$

Step 2. Calculate the mass of $\text{NaCl}$

$\text{Mass of NaCl" = 1025 color(red)(cancel(color(black)("g seawater"))) × "3.50 g NaCl"/(100 color(red)(cancel(color(black)("g seawater")))) = "35.88 g NaCl}$

Step 3. Calculate the moles of $\text{NaCl}$

$\text{Moles of NaCl" = 35.88 color(red)(cancel(color(black)("g NaCl"))) × "1 mol NaCl"/(58.44 color(red)(cancel(color(black)("g NaCl")))) = "0.6139 mol NaCl}$

Step 4. Calculate the molarity of $\text{NaCl}$

$\text{ Molarity" = "0.6139 mol"/"1 L" = "0.614 mol/L}$