The mass of sodium acetate () required to prepare 250 mL of 0.35 M aqueous solution is _____ g.

Question

The mass of sodium acetate ( ${CH}_{3} COONa$ ) required to prepare 250 mL of 0.35 M aqueous solution is _____ g.
(Molar mass of ${CH}_{3} COONa$ is 82.02 g ${mol}^{- 1}$ ) [2024]

Smart Achievers · Accepted Answer

(7)

$Volume of solution (V) = 250 mL = 0.25 L .$

${Moles of CH}_{3} COONa (n)$

$= Molarity \times volume = 0.35 \frac{mol}{L} \times 0.25 L = 0.0875 mol$

${Mass of CH}_{3} COONa$

$= n \times molar mass = 0.0875 mol \times 82.02 \frac{g}{mol} = 7.177 g$

Solution

Q.
The mass of sodium acetate ( ${CH}_{3} COONa$ ) required to prepare 250 mL of 0.35 M aqueous solution is _____ g.
(Molar mass of ${CH}_{3} COONa$ is 82.02 g ${mol}^{- 1}$ ) [2024]

Ans.
(7)

$Volume of solution (V) = 250 mL = 0.25 L .$

${Moles of CH}_{3} COONa (n)$

$= Molarity \times volume = 0.35 \frac{mol}{L} \times 0.25 L = 0.0875 mol$

${Mass of CH}_{3} COONa$

$= n \times molar mass = 0.0875 mol \times 82.02 \frac{g}{mol} = 7.177 g$

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Solution

Q. 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-1) [2024]

Ans. (7) Volume of solution (V)=250 mL=0.25 L. Moles of CH3COONa (n) =Molarity×volume=0.35 molL×0.25 L=0.0875 mol Mass of CH3COONa =n×molar mass=0.0875 mol×82.02 gmol=7.177 g

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Q.
The mass of sodium acetate ( ${CH}_{3} COONa$ ) required to prepare 250 mL of 0.35 M aqueous solution is _____ g.
(Molar mass of ${CH}_{3} COONa$ is 82.02 g ${mol}^{- 1}$ ) [2024]

Ans.
(7)

$Volume of solution (V) = 250 mL = 0.25 L .$

${Moles of CH}_{3} COONa (n)$

$= Molarity \times volume = 0.35 \frac{mol}{L} \times 0.25 L = 0.0875 mol$

${Mass of CH}_{3} COONa$

$= n \times molar mass = 0.0875 mol \times 82.02 \frac{g}{mol} = 7.177 g$