If 23 g of ethanol (C_2H_5OH,molar mass = 46 | vedclass

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(C) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.First,calculate the number of moles of ethanol $(C_2H_5OH)$:$	ex

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(C) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.First,calculate the number of moles of ethanol $(C_2H_5OH)$:$	ext{Moles} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{23 \ g}{46 \ g \ mol^{-1}} = 0.5 \ mol$.Next,convert the volume of the solution from $mL$ to $L$:$500 \ mL = 0.5 \ L$.Finally,calculate the molarity:$M = \frac{	ext{moles of solute}}{	ext{volume of solution in } L} = \frac{0.5 \ mol}{0.5 \ L} = 1 \ M$.

If $23 \ g$ of ethanol ($C_2H_5OH$,molar mass $= 46 \ g \ mol^{-1}$) is dissolved in $500 \ mL$ of solution,find the molarity of the solution in $M$.

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