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 $litr{e}^{-1}$ $se{c}^{-1}$, how long would it take for $H^{+}$ in the drop to disappear?

• A. $6\times {10}^{-8}$ sec
• B. $6\times {10}^{-9}$ sec
• C. $6\times {10}^{-10}$ sec
• D. $6\times {10}^{-12}$ sec

Answer 1

The correct option is B $6\times {10}^{-9}$ sec

Let us consider the problem:-

rate constant $=1.0\times {10}^{7}mollitr{e}^{-1}{\sec }^{-1}\left(Given\right)$

Using Zero Order $t=\frac{x}{k}$ (1)

where, $x$ is Concentration used $k$ rate constant

Therefore,

$0.05mlhas\frac{3\times {10}^{-6}\times {10}^3}{0.05}=0.06\times {10}^{-1}mol/litre$

Using equation (1), we have

$t=\frac{0.6\times {10}^{-1}}{1\times {10}^7}=6\times {10}^{-9}$