A drop of sol...

Question

A drop of solution(volume $0.05$ mL) contains $3 \times 10^{- 6}$ mole of $H^{+}$ . If the rate constant of disappearance of $H^{+}$ is $10^{7}$ mol $l i t r e^{- 1}$ $s e c^{- 1}$ , how long would it take for $H^{+}$ in the drop to disappear?

6 \times 10^{- 8}

sec

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6 \times 10^{- 9}

sec

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6 \times 10^{- 10}

sec

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6 \times 10^{- 12}

sec

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Solution

The correct option is B $6 \times 10^{- 9}$ sec
Let us consider the problem:-
rate constant $= 1.0 \times 10^{7} m o l l i t r e^{- 1} {sec}^{- 1} (G i v e n)$
Using Zero Order $t = \frac{x}{k}$ (1)
where, $x$ is Concentration used $k$ rate constant
Therefore,
$0.05 m l h a s \frac{3 \times 10^{- 6} \times 10^{3}}{0.05} = 0.06 \times 10^{- 1} m o l / l i t r e$
Using equation (1), we have
$t = \frac{0.6 \times 10^{- 1}}{1 \times 10^{7}} = 6 \times 10^{- 9}$

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