$10 m L$ of $2 (M) N a O H$ solution is added to $200 m L$ of $0.5 (M)$ of $N a O H$ solution. What is the final concentration?

0.3 M

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0.6 M

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0.5 M

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0.4 M

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Solution

The correct option is B 0.6 M
Number of moles of $N a O H$ in 10 mL of 2 M solution $= \frac{10 m L}{1000 m L / L} \times 2 m o l / L = 0.02 m o l$
Number of moles of $N a O H$ in 200 mL of 0.5 M solution $= \frac{200 m L}{1000 m L / L} \times 0.5 m o l / L = 0.1 m o l$
Total number of moles of $N a O H = 0.02 + 0.1 = 0.12 m o l$
Total volume $= 10 + 200 = 210 m L = \frac{210 m L}{1000 m L / L} = 0.210 L$
Final concentration $= \frac{0.12 m o l}{0.210 L} = 0.57 M$

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