What volume of water is to be added to $100 c m^{3}$ of $0.5 M N a O H$ solution to make it $0.1 M$ solution?

200 c m^{3}

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400 c m^{3}

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500 c m^{3}

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100 c m^{3}

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Solution

The correct option is B $400 c m^{3}$
Initial molarity, $M_{1} = 0.5 M$ , initial volume, $V_{1} = 100 c m^{3}$
Final molarity, $M_{2} = 0.1 M$ , initial volume, $V_{1} = V c m^{3}$
using $M_{1} V_{1} = M_{2} V_{2}$
$0.5 \times 100 = 0.1 \times V$
$V = 500 c m^{3}$
Volume of water added $= 500 - 100 = 400 c m^{3}$

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