A drop of solution (volume 0.05 mL) contains $3 .0 \times 10^{- 6}$ moles of $H^{+}$ . If the rate constant of disappearance of $H^{+}$ is $1 .0 \times 10^{7} mole {litre}^{- 1} \sec^{- 1}$ . How long would it take for $H^{+}$ in drop to disappear:

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$6 \times 10^{- 8} \sec$

$6 \times 10^{- 7} \sec$

$6 \times 10^{- 9} \sec$

$6 \times 10^{- 10} \sec$

answer is C.

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Detailed Solution

Units rate constant $= {mol L}^{- 1} \sec^{- 1} \Rightarrow$ Zero order reaction.

$[H^{+}] = \frac{3 \times 10^{- 6} \times 1000}{0.05} = 0.06 mol . L^{- 1} t = \frac{x}{K} = \frac{6 \times 10^{- 2}}{10^{7}} = 6 \times 10^{- 9} \sec$

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