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$10 \mathrm{~mL}$ of $2(\mathrm{M}) \mathrm{~NaOH}$ solution is added to $200 \mathrm{~mL}$ of $0.5$ $\mathrm{(M)}$ of $\mathrm{NaOH}$ solution. What is the final concentration ?
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Verified Answer
The correct answer is:
$0.57(\mathrm{M})$
$0.57(\mathrm{M})$
From molarity equation
$\begin{aligned}
& \mathrm{M}_1 \mathrm{V}_1+\mathrm{M}_2 \mathrm{V}_2=\mathrm{MV}_{\text {(total) }} \\
& 2 \times \frac{10}{1000}+0.5 \times \frac{200}{1000}=\mathrm{M} \times \frac{210}{1000} \\
& 120=\mathrm{M} \times 210 \\
& \mathrm{M}=\frac{120}{210}=0.57 \mathrm{~M}
\end{aligned}$
$\begin{aligned}
& \mathrm{M}_1 \mathrm{V}_1+\mathrm{M}_2 \mathrm{V}_2=\mathrm{MV}_{\text {(total) }} \\
& 2 \times \frac{10}{1000}+0.5 \times \frac{200}{1000}=\mathrm{M} \times \frac{210}{1000} \\
& 120=\mathrm{M} \times 210 \\
& \mathrm{M}=\frac{120}{210}=0.57 \mathrm{~M}
\end{aligned}$
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