Question:
The mass of sodium acetate required to prepare $250$ mL of $0.4$ M aqueous solution is \
(Molar mass of sodium acetate = $82$ g mol$^{-1}$)
The mass of sodium acetate required to prepare $250$ mL of $0.4$ M aqueous solution is \
(Molar mass of sodium acetate = $82$ g mol$^{-1}$)
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For solution questions, use directly: Mass = Molarity × Volume(in L) × Molar Mass.
Updated On: Apr 24, 2026
- $4.1$ g
- $8.2$ g
- $16.4$ g
- $2.05$ g
- $0.82$ g
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The Correct Option is B
Solution and Explanation
Concept:
Chemistry - Molarity.
[
M=\frac{\text{moles of solute}}{\text{volume in litre}}
]
Step 1: Convert volume into litre. [ 250\text{ mL}=0.250\text{ L} ]
Step 2: Find moles required. [ \text{Moles}=M \times V=0.4 \times 0.250=0.1 ]
Step 3: Find mass of sodium acetate. [ \text{Mass}=\text{moles} \times \text{molar mass} ] [ =0.1 \times 82=8.2\text{ g} ]
Step 4: Final answer. [ \boxed{8.2\text{ g}} ]
Hence, correct option is (B).
Step 1: Convert volume into litre. [ 250\text{ mL}=0.250\text{ L} ]
Step 2: Find moles required. [ \text{Moles}=M \times V=0.4 \times 0.250=0.1 ]
Step 3: Find mass of sodium acetate. [ \text{Mass}=\text{moles} \times \text{molar mass} ] [ =0.1 \times 82=8.2\text{ g} ]
Step 4: Final answer. [ \boxed{8.2\text{ g}} ]
Hence, correct option is (B).
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