Question

We have 32.4 mg NaCl in 122.4 mL of solution. The molarity of this solution is

a-32.4 M NaCl
b-0.554 M NaCl
c-0.00453 M NaCl
d-0.1224 M NaCl

Answer 1

The molarity of the solution is 0.00453 M NaCl.

Step-by-step explanation:

The molarity of a solution can be calculated using the formula:

Molarity (M) = moles of solute / volume of solution (L)

Given that we have 32.4 mg (0.0324 g) of NaCl in 122.4 mL (0.1224 L) of solution, we need to convert the mass of NaCl to moles and the volume of solution to liters:

1. Convert mass of NaCl to moles using the molar mass of NaCl (58.44 g/mol): 0.0324 g / 58.44 g/mol = 0.000554 mol NaCl
2. Convert volume of solution to liters: 0.1224 L

Now, we can substitute these values into the formula to calculate the molarity:

Molarity (M) = 0.000554 mol / 0.1224 L = 0.00453 M NaCl

Therefore, the correct answer is c) 0.00453 M NaCl.