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3.0 molal NaOH solution has a density of 1.110g/mL. The molarity of the solution is:

A. 2.9732
B. 3.05
C. 3.64
D. 3.0504

User Graham Lea
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Final answer:

To find the molarity of a 3.0 molal NaOH solution with a density of 1.110 g/mL, calculate the mass of NaOH (120 g) in one kilogram of water, convert this to the volume of the solution (1009 mL), then divide the moles of NaOH by this volume to find the molarity, which is approximately 2.97 M.

Step-by-step explanation:

To calculate the molarity of a 3.0 molal NaOH solution with a density of 1.110 g/mL, we first need to determine the mass of the solvent (water) in 1 liter of the solution. Since the solution is 3.0 molal, this means there are 3.0 moles of NaOH per kilogram of water. We'll use the molar mass of NaOH (40.00 g/mol) to find the mass of NaOH and then the total mass of the solution.

Assuming 1 kg of water is present, the mass of NaOH is:

3.0 mol × 40.00 g/mol = 120.0 g of NaOH

So, the mass of the solution is:

1000 g (water) + 120 g (NaOH) = 1120 g of solution

Now, we'll convert the mass of the solution to volume using the density:

1120 g ÷ 1.110 g/mL = 1009 mL or 1.009 L

The molarity (M) is the number of moles of solute (NaOH) per liter of solution:

3.0 mol NaOH ÷ 1.009 L = 2.97 M NaOH

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User Arief Hidayatulloh
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