The mass of sodium acetate (CH_3COONa) requir | vedclass

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(B) The number of moles of solute is calculated as: $Moles = Molarity 	imes Volume \ (L)$.Given $Molarity = 0.35 \ M$ and $Volume = 250 \ mL = 0.25 \

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The number of moles of solute is calculated as: $Moles = Molarity 	imes Volume \ (L)$.Given $Molarity = 0.35 \ M$ and $Volume = 250 \ mL = 0.25 \ L$.$Moles = 0.35 \ mol \ L^{-1} 	imes 0.25 \ L = 0.0875 \ mol$.The mass of the solute is calculated as: $Mass = Moles 	imes Molar \ mass$.$Mass = 0.0875 \ mol 	imes 82.02 \ g \ mol^{-1} = 7.17675 \ g \approx 7.18 \ g$.Rounding to the nearest integer as per the options,the correct value is $7 \ g$.

The mass of sodium acetate $(CH_3COONa)$ required to prepare $250 \ mL$ of $0.35 \ M$ aqueous solution is . . . . . . $g$. (Molar mass of $CH_3COONa$ is $82.02 \ g \ mol^{-1}$)

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