Question

a 2.61 l aqueous solution of koh contains 199 g of koh . the solution has a density of 1.29 g/ml . calculate the molarity ( ), molality ( ), and mass percent concentration of the solution?

Answer 1

The molarity of the KOH solution is 1.359 M, the molality is 1.121 m, and the mass percent concentration is 5.92%.

Step-by-step explanation:

To find the molarity, molality, and mass percent concentration of the KOH solution, we'll need to follow several steps. First, molarity (M) is moles of solute per liter of solution. Molality (m) is moles of solute per kilogram of solvent. Mass percent is mass of solute divided by total mass of solution times 100.

For molarity, we will use the formula:

Molarity (M) = moles of KOH / volume of solution in liters

Molar mass of KOH = 39.1 (K) + 16 (O) + 1.01 (H)

= 56.11 g/mol

Moles of KOH = 199 g / 56.11 g/mol

= 3.548 moles

Volume of solution = 2.61 L

M = 3.548 moles / 2.61 L

= 1.359 M

For molality, the formula is:

Molality (m) = moles of solute / mass of solvent in kg

Mass of solution = volume × density

Mass of solution = 2.61 L × 1.29 g/mL

= 3364.9 g

Mass of solvent = mass of solution - mass of solute = 3364.9 g - 199 g

= 3165.9 g

= 3.1659 kg

m = 3.548 moles / 3.1659 kg

= 1.121 m

The mass percent concentration is calculated as follows:

Mass percent = (mass of solute / mass of solution) × 100

Mass percent = (199 g / 3364.9 g) × 100 =

5.92%