The amount of water that should be added to $500 m L$ of $0.5 N$ solution of $N a O H$ to give a concentration of $10 m g$ per mL is

100

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200

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250

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500

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Solution

The correct option is D $500$
$N_{1} = 0.5 N \to 10 m g p e r m L$
$N_{2} = \frac{10 \times 10^{- 3} g m}{40 \times 1} \times 1000 = 0.25 N$
$V_{1} = 500 m l, V_{2} = ?$
$N_{1} V_{1} = N_{2} V_{2}; 0.5 \times 500 = 0.25 \times V_{2}$
$V_{2} = 1000 m L$ final volume water added $= 1000 - 500 = 500 m L$

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