Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

To determine the amount of benzoic acid (C6H5COOH) needed to prepare a 250 mL solution with a concentration of 0.15 M in methanol, we can follow a straightforward calculation process. Molarity (M) is defined as the number of moles of solute per liter of solution. Therefore, we can use the formula:

Calculating Moles of Solute

The formula for molarity is:

Molarity (M) = Moles of solute / Volume of solution in liters

From this, we can rearrange the formula to find the number of moles of solute required:

Moles of solute = Molarity (M) × Volume of solution (L)

Step 1: Convert Volume to Liters

First, we need to convert the volume of the solution from milliliters to liters:

• 250 mL = 250 / 1000 = 0.250 L

Step 2: Calculate Moles of Benzoic Acid

Now, we can substitute the values into the rearranged formula:

Moles of benzoic acid = 0.15 M × 0.250 L

Calculating this gives:

Moles of benzoic acid = 0.0375 moles

Finding the Mass of Benzoic Acid

Next, we need to convert moles of benzoic acid into grams. To do this, we need the molar mass of benzoic acid. The molar mass can be calculated by adding the atomic masses of all the atoms in the molecular formula C6H5COOH:

• Carbon (C): 12.01 g/mol × 7 = 84.07 g/mol
• Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol
• Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol

Adding these together gives:

Molar mass of C6H5COOH = 84.07 + 6.048 + 32.00 = 122.118 g/mol

Step 3: Calculate the Mass Required

Now, we can find the mass of benzoic acid needed using the formula:

Mass (g) = Moles × Molar Mass

Substituting the values we have:

Mass = 0.0375 moles × 122.118 g/mol

Calculating this gives:

Mass = 4.58 g

Final Result

To prepare 250 mL of a 0.15 M solution of benzoic acid in methanol, you will need approximately 4.58 grams of benzoic acid. Make sure to measure accurately and mix thoroughly to ensure the solution is homogeneous.

To find out how much benzoic acid (C6H5COOH) you need to prepare a 250 mL solution at a concentration of 0.15 M, we can use the formula for molarity, which is defined as the number of moles of solute divided by the volume of solution in liters. Let's break this down step by step.

Step 1: Calculate the number of moles required

The first thing we need to do is calculate the number of moles of benzoic acid needed for the solution. The formula for molarity (M) is:

M = n / V

Where:

• M is the molarity (in moles per liter)
• n is the number of moles of solute
• V is the volume of the solution in liters

We know the molarity (0.15 M) and the volume (250 mL), but we need to convert the volume from milliliters to liters:

V = 250 mL = 250 / 1000 = 0.250 L

Now we can rearrange the formula to solve for n (the number of moles):

n = M × V

Substituting the values we have:

n = 0.15 mol/L × 0.250 L = 0.0375 moles

Step 2: Calculate the mass of benzoic acid needed

Next, we need to convert the number of moles into grams. To do this, we need the molar mass of benzoic acid. The molecular formula is C6H5COOH, and we can calculate its molar mass as follows:

• Carbon (C): 12.01 g/mol × 7 = 84.07 g/mol
• Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol
• Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol

Now, adding these together gives us:

Molar mass of C6H5COOH = 84.07 + 6.048 + 32.00 = 122.12 g/mol

Now we can find the mass of benzoic acid needed using the formula:

mass = n × molar mass

Substituting the values:

mass = 0.0375 moles × 122.12 g/mol = 4.58 grams

Final Calculation

To prepare a 250 mL solution of 0.15 M benzoic acid in methanol, you will need approximately 4.58 grams of benzoic acid. Make sure to weigh it accurately and dissolve it in methanol to achieve the desired concentration.

This method can be applied to any solute and solvent combination, as long as you know the molarity you want to achieve and the volume of the solution you plan to prepare. Understanding these calculations is essential for accurate solution preparation in chemistry.