A 500 , mL solution of NaOH contains 3.01 tim | vedclass

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(A) Step $1$: Calculate the number of moles of $NaOH$. Number of moles = $\frac{	ext{Number of molecules}}{	ext{Avogadro's number}} = \frac{3.01 	i

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(A) Step $1$: Calculate the number of moles of $NaOH$. Number of moles = $\frac{	ext{Number of molecules}}{	ext{Avogadro's number}} = \frac{3.01 	imes 10^{22}}{6.022 	imes 10^{23}} \approx 0.05 \, mol$. Step $2$: Convert volume to liters. Volume = $500 \, mL = 0.5 \, L$. Step $3$: Calculate Molarity $(M)$. $M = \frac{	ext{moles of solute}}{	ext{volume of solution in liters}} = \frac{0.05 \, mol}{0.5 \, L} = 0.1 \, M$.

$A$ $500 \, mL$ solution of $NaOH$ contains $3.01 \times 10^{22}$ molecules of $NaOH$. What is the molarity $(M)$ of the $NaOH$ solution?

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