# How many grams of solute are present in 1.6 L of 3.0 M NaOH?

## Calculating the Mass of Solute in a Solution

In chemistry, it is important to be able to determine the amount of solute present in a given volume of solution. One common way to do this is by using the formula:

**mass = molarity × volume × molar mass**

Let's apply this formula to the scenario presented in the question:

## Options:

a) 4.8 g

b) 14.4 g

c) 9.6 g

d) 1.2 g

### Final answer:

To find the mass of solute in 1.6 L of 3.0 M NaOH, multiply the molarity (3.0 M) by the volume (1.6 L) to get the moles of NaOH, and then multiply by the molar mass of NaOH (40.0 g/mol) to get 192 g.

### Explanation:

To calculate the mass of solute present in 1.6 L of 3.0 M NaOH, we will follow these steps:

- Identify the molarity (M) of the solution, which is given as 3.0 M NaOH.
- Use the volume of the solution, which is 1.6 L.
- Multiply the molarity by the volume to find the number of moles of NaOH. This gives us 3.0 moles/L × 1.6 L = 4.8 moles NaOH.
- Now, convert the amount of solute to grams, using the molar mass of NaOH, which is 40.0 g/mol: 4.8 moles × 40.0 g/mol = 192 g of NaOH.

Therefore, the solution contains 192 g of NaOH, which is not an option listed above. It appears there may be a mistake in the question or the provided options.

What is the final mass of solute in the solution? The final mass of solute in 1.6 L of 3.0 M NaOH is 192 grams.