Question

A drop of a solution (volume $=0.05ml$) contains $6\times {10}^{-7}mole$ of $H^{+}$ .

If the rate of disappearance of $H^{+}$ is $6.0\times {10}^{5}mol{L}^{-1}{s}^{-1}$.

How long will it take for the $H^{+}$ in the drop to disappear ?

• A. $8.0\times {10}^{-8}$ s
• B. $2.0\times {10}^{-8}$ s
• C. $6.0\times {10}^{-6}$ s
• D. $2.0\times {10}^{-2}$ s

Answer 1

The correct option is B $2.0\times {10}^{-8}$ s

$6\times {10}^{-7}$ moles in $0.05ml$ corresponds to $\frac{6\times {10}^{-7}moles}{\frac{0.05}{1000}}$

$=1.2\times {10}^{-2}moles.$

In one second, $6.0\times {10}^5$ moles disappear.

Hence, the time required for the disappearance of $1.2\times {10}^{-2}$ moles is $\frac{1.2\times {10}^{-2}}{6.0\times {10}^5}$

$=2\times {10}^{-8}sec.$