A sample of seawater contains 78.0 grams of NaCl in 2,025 mL of solution. If the molar mass of NaCl is 58.443 g/mol, what is the molarity of this sample? 0.659 M NaCl
0.823 M NaCl
1.84 M NaCl
2.08 M NaCl

that would be 0.659 M NaCl

0.659 M NaCl

Explanation:

Equation: Molarity = moles of solute / liters of solution

1. To find moles of solute:

78.0 g NaCl x 1 mol NaCl / 58.44 g NaCl

= 1.334 mol NaCl

2. Calculate molarity

Transfer 2,025 mL to L

1.334 / 2.025 = 0.659 M NaCl (rounded)

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

the answer to your question is c