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QJEEM24Ch7e3b: JEE Main 2024 Chemistry - Class 11 JEEM24Ch7e3b Solution: The mass of sodium acetate required to prepare of aqueous…

The mass of sodium acetate (CH3COONa)\left(\mathrm{CH}_3 \mathrm{COONa}\right) required to prepare 250 mL250 \mathrm{~mL} of 0.35 M0.35 \mathrm{~M} aqueous solution is ________ g. (Molar mass of CH3COONa\mathrm{CH}_3 \mathrm{COONa} is 82.02 g mol−182.02 \mathrm{~g} \mathrm{~mol}^{-1} )

Step-by-Step Verified Solution

amp; Moles = Molarity × Volume in litres amp;=0.35×0.25amp; Mass = moles × molar mass amp;=0.35×0.25×82.02=7.18 g\begin{aligned}& \text { Moles }=\text { Molarity } \times \text { Volume in litres } \\& =0.35 \times 0.25 \\& \text { Mass }=\text { moles } \times \text { molar mass } \\& =0.35 \times 0.25 \times 82.02=7.18 \mathrm{~g}\end{aligned}

Ans. 7

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