What weight of NaOH would be required to prepare 3000ml of a 2.5M solution?

M= moles/L

mols= grams/molar mass

300 grams

To determine the weight of NaOH required to prepare a 3000 ml of 2.5M solution, you need to follow these steps:

Step 1: Understand the formula for molarity (M) and its units.
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula for molarity is:

M = moles of solute / volume of solution (in liters)

Step 2: Convert the volume of the solution to liters.
Since the given volume is in milliliters (ml), you need to convert it to liters. There are 1000 milliliters in a liter, so the volume of the solution in liters would be:
3000 ml ÷ 1000 = 3 liters

Step 3: Use the molarity formula to find the number of moles of NaOH.
Given Molarity = 2.5M and Volume of solution = 3 liters, you can rearrange the formula to find the moles of solute:
Moles of solute = Molarity × Volume of solution

Moles of solute = 2.5M × 3L = 7.5 moles

Step 4: Use the molar mass of NaOH to convert moles to grams.
The molar mass of NaOH is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H). It can be calculated using the periodic table. The molar mass of NaOH is approximately 40 g/mol.

To calculate the weight (mass) of NaOH required, multiply the moles of NaOH by its molar mass:
Weight of NaOH = Moles of NaOH × Molar mass of NaOH

Weight of NaOH = 7.5 moles × 40 g/mol = 300 grams

Therefore, approximately 300 grams of NaOH would be required to prepare 3000 ml of a 2.5M solution.