# Question #6469b

##### 1 Answer

The *molality* of the solution is **2.29 molal**.

#### Explanation:

There are a couple of things wrong with your question

- you know that the solution's
*molarity*is**2.05 M**, so I presume that you're interested in its*molality*; - the density of acetic acid solutions doesn't come close to 1.20 g/mL (
*pure acetic acid*has a density of 1.05 g/mL), so I presume that you meant to write**1.02 g/mL**.

This being said, you need to focus on how to get the solution's molality, which is defined as moles of solute, in your case acetic acid, per **kilograms** of solvent, in your case water.

Let's assume a **1-L** sample of your solution. This solution would contain

In order to get the mass of water, you need to use the solution's density.

Since the solution contains **2.05 moles** of acetic acid, you can use its molar mass to determine how many grams of acetic acid you have.

This measn that the solution will contain

This means that the solution's molality will be

Rounded to three sig figs, the answer will be