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What is the molarity of a solution that contains 238.0 g NaOH in 9.3 L solution?

a 1.0 M
b 2.0 M
c 3.0 M
d 4.0 M

1 Answer

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Final answer:

The molarity of a solution containing 238.0 g NaOH in 9.3 L is calculated to be 0.64 M by first converting the mass of NaOH to moles using its molar mass and then dividing by the volume of the solution.

Step-by-step explanation:

To find the molarity of a solution containing 238.0 g NaOH in 9.3 L, we first need to convert the mass of NaOH to moles. The molar mass of NaOH is 40.0 g/mol. So, we calculate the moles of NaOH by dividing the mass by the molar mass:

238.0 g NaOH ÷ (40.0 g NaOH/mol) = 5.95 mol NaOH

Next, we use the definition of molarity, which is moles of solute per liter of solution, to find the molarity:

5.95 mol NaOH ÷ 9.3 L = 0.64 M NaOH

So, the molarity of the solution is 0.64 M, which isn't one of the options provided (1.0 M, 2.0 M, 3.0 M, or 4.0 M). There may have been either a mistake in the choices provided or a misinterpretation in the question.

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