Question

The mass of sodium acetate (CH₃COONa) required to prepare 250 mL of 0.35 M aqueous solution is g. (Molar mass of CH₃COONa is 82.02 g/mol)

Answer 1

Correct Answer: 7

First Calculate the moles using the formula:
Moles = Molarity × Volume in litres
Moles\(= 0.35 × 0.25=0.0875 mol\)

Now Calculate the mass of sodium acetate:
Mass = moles × molar mass
Mass\(= 0.35 × 0.25 × 82.02 = 7.18 g≈ 7 g\)

Answer 2

The problem asks to calculate the mass of sodium acetate (\(\text{CH}_3\text{COONa}\)) required to prepare 250 mL of a 0.35 M aqueous solution, given the molar mass of sodium acetate.

Concept Used:

The solution is based on the definition of Molarity, which is a measure of the concentration of a solute in a solution. Molarity (M) is defined as the number of moles of solute per liter of solution.

The key formulas are:

1. Molarity formula: \[ M = \frac{\text{Moles of solute}}{\text{Volume of solution (in L)}} \]
2. Relationship between mass, moles, and molar mass: \[ \text{Moles of solute} = \frac{\text{Mass of solute}}{\text{Molar mass of solute}} \]

By combining these two formulas, we can directly solve for the required mass of the solute.

Step-by-Step Solution:

Step 1: List the given information and convert the volume to Liters (L).

• Molarity of the solution, \(M = 0.35 \, \text{M} = 0.35 \, \text{mol/L}\)
• Volume of the solution, \(V = 250 \, \text{mL}\)
• Molar mass of \(\text{CH}_3\text{COONa}\), \(M_w = 82.02 \, \text{g/mol}\)

Convert the volume from mL to L:

\[ V = 250 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.250 \, \text{L} \]

Step 2: Calculate the number of moles of \(\text{CH}_3\text{COONa}\) required.

Rearranging the molarity formula, we get:

\[ \text{Moles of solute} = M \times V (\text{in L}) \]

Substituting the given values:

\[ \text{Moles} = 0.35 \, \text{mol/L} \times 0.250 \, \text{L} \] \[ \text{Moles} = 0.0875 \, \text{mol} \]

Step 3: Calculate the mass of \(\text{CH}_3\text{COONa}\) required.

Using the relationship between mass, moles, and molar mass:

\[ \text{Mass of solute} = \text{Moles} \times \text{Molar mass} \]

Substituting the values from Step 2 and the given molar mass:

\[ \text{Mass} = 0.0875 \, \text{mol} \times 82.02 \, \text{g/mol} \]

Final Computation & Result:

Performing the final multiplication:

\[ \text{Mass} = 7.17675 \, \text{g} \]

Rounding to two decimal places for practical purposes, the required mass is 7.18 g.

The mass of sodium acetate required is 7.18 g.