$4 g m$ of $N a O H$ are dissolved in $500 m l$ of water. Active mass of $N a O H$ is :

0.1

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0.2

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0.3

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0.4

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Solution

The correct option is B $0.2$
Active mass: Concentration of a substance expressed usually in moles per liter. (Molarity basically)

Number of moles $= \frac{4}{40} = 0.1 m o l e s$

$V = 500 m l = 0.5 l i t r e s$

Molarity / Active mass $= \frac{0.1}{0.5} = 0.2 m o l / l i t r e$

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