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)
Correct Answer: 7
Approach Solution - 1
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\(= 0.35 × 0.25 × 82.02 = 7.18 g≈ 7 g\)
Approach Solution -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:
- Molarity formula: \[ M = \frac{\text{Moles of solute}}{\text{Volume of solution (in L)}} \]
- 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.
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