10 mL of 2 M NaOH solution is added to 20 | vedclass

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(B) Step $1$: Reaction between $NaOH$ and $HCl$ in the beaker: $NaOH + HCl ightarrow NaCl + H_2O$ Initial moles of $NaOH = 2 \ M 	imes 0.01 \ L = 0

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$20 \ M$ $HCl$ solution

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(B) Step $1$: Reaction between $NaOH$ and $HCl$ in the beaker: $NaOH + HCl ightarrow NaCl + H_2O$ Initial moles of $NaOH = 2 \ M 	imes 0.01 \ L = 0.02 \ mol$. Initial moles of $HCl = 1 \ M 	imes 0.02 \ L = 0.02 \ mol$. Since the moles are equal,the resulting solution is neutral ($NaCl$ solution). Step $2$: $10 \ mL$ of this neutral solution is added to a flask containing $2 \ moles$ of $HCl$. The total volume of the flask is made up to $100 \ mL$ $(0.1 \ L)$. Molarity of $HCl = \frac{	ext{moles of } HCl}{	ext{Volume of solution in } L} = \frac{2 \ mol}{0.1 \ L} = 20 \ M$.

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