Question

What mass of ch3oh is required to prepare 109.1 ml of solution with a ch3oh concentration of 0.342 m?

A) 2.75 g
B) 5.50 g
C) 8.25 g
D) 11.00 g

Answer 1

To find the mass of CH3OH required, convert the volume to liters, calculate the moles using the molarity and volume, and then use the molar mass of CH3OH to find the mass. To prepare the solution, 1.18 g of CH3OH is required. The correct answer is D) 1.18 g.

Step-by-step explanation:

To calculate the mass of CH3OH (methanol) necessary to prepare a 109.1 mL solution with a concentration of 0.342 M, we can use the formula related to molarity:

Molarity (M) = moles of solute / liters of solution

Step 1: Convert volume from mL to L:

109.1 mL * (1 L / 1000 mL) = 0.1091 L

Step 2: Calculate the moles of CH3OH:

0.342 M * 0.1091 L = moles of CH3OH

Step 3: Calculate the mass of CH3OH using its molar mass (32.04 g/mol):

Mass = moles * molar mass

By following these steps, we can determine which of the given options (A, B, C, D) corresponds to the correct mass required for the solution.