.35 \. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,NCERT,NEET,Solutions,Concentration Terms, Class 11 Chemistry,Class 12 Chemistry,JEE Main Chemistry.">
.35 \. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,NCERT,NEET,Solutions,Concentration Terms, Class 11 Chemistry,Class 12 Chemistry,JEE Main Chemistry.">
.35 \. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,NCERT,NEET,Solutions,Concentration Terms, Class 11 Chemistry,Class 12 Chemistry,JEE Main Chemistry.">
The mass of sodium acetate $\left(\mathrm{CH}_3 \mathrm{COONa}\right)$ required to prepare $250 \mathrm{~mL}$ of $0.35 \mathrm{~M}$ aqueous solution is ________ g. (Molar mass of $\mathrm{CH}_3 \mathrm{COONa}$ is $82.02 \mathrm{~g} \mathrm{~mol}^{-1}$)
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Step-by-Step Solution
Step 1: Convert Volume to Liters
The given volume of the solution is 250 mL. To use it in a molarity calculation, first convert it to liters:
\text{Volume (in liters)} = \frac{250}{1000} = 0.25 \,\text{L}
Step 2: Calculate Moles of Sodium Acetate
Molarity (M) is defined as the number of moles of solute per liter of solution. Thus:
\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
From this equation, the moles of sodium acetate required are:
\text{moles of sodium acetate} = 0.35 \,\text{M} \times 0.25 \,\text{L} = 0.0875 \,\text{mol}
Step 3: Calculate Mass of Sodium Acetate
To find the mass, multiply the number of moles of sodium acetate by its molar mass ( 82.02 \,\text{g mol}^{-1} ):
\text{Mass of sodium acetate} = 0.0875 \,\text{mol} \times 82.02 \,\text{g mol}^{-1} = 7.18 \,\text{g}
Step 4: State the Final Answer
Therefore, the required mass of sodium acetate to prepare 250 mL of a 0.35 M solution is
7.18 g.