A solution is made by dissolving 42.3 g of potassium hydroxide
(KOH) in 329 g of water.
What is the molality of the solution?

2.29 m.

Explanation:

The following data were obtained from the question:

Mass of KOH = 42.3 g

Molar mass of KOH = 56.11 g/mol

Mass of water = 329 g

Molality of KOH = ?

Next, we shall determine the number of mole in 42.3 g of KOH. This can be obtained as follow:

Mass of KOH = 42.3 g

Molar mass of KOH = 56.11 g/mol

Mole of KOH =?

Mole = mass /Molar mass

Mole of KOH = 42.3/56.11

Mole of KOH = 0.754 mole

Next, we shall convert 329 g of water to kilogram (kg). This can be obtained as follow:

1000 g = 1 kg

Therefore,

329 g = 329 g /1000 g × 1 kg

329 g = 0.329 kg

Therefore, 329 g of water is equivalent to 0.329 kg

Finally, we shall determine the molality of the KOH solution ad follow:

Molality is defined as the mole of solute per unit kilogram of solvent (water) i.e

Molality = mole/ mass (kg) of water

Mole of KOH = 0.754 mole.

Mass of water = 0.329 kg.

Molality = mole/ mass (kg) of water

Molality = 0.754/0.329

Molality = 2.29 m

Therefore, the molality of the KOH solution is 2.29 m.

the answer is definitely c.