A drop of a solution $\left( {{\rm{volume}} = 0.05\;{\rm{mL}}} \right)$ contains $6 \times {10^{ - 7}}\;mol\;of\;{{\rm{H}}^ + }.$ If the rate of disappearance of ${{\rm{H}}^ + }\;is\;6.0 \times {10^5}\;{\rm{mol}}/{\rm{L}} \times {\rm{s}},$ how long will it take for ${{\rm{H}}^ + }$ to disappear from the drop

$2.0 \times {10^{ - 8}}\;{\rm{s}}$

$8.0 \times {10^{ - 8}}\;{\rm{s}}$

$6.0 \times {10^{ - 6\;}}{\rm{s}}$

$2.0 \times {10^{ - 2}}\;{\rm{s}}$

Back to CHEMICAL KINETICS

NEET UG CHEMISTRY MCQ: CHEMICAL KINETICS

1 Hard

1 of 1

Application +4 | -1

A drop of a solution

\left( {\mathrm{volume} = 0.05\;\mathrm{mL}} \right)

contains

6 \times {10^{ - 7}}\;mol\;of\;{\mathrm{H}^ + }.

If the rate of disappearance of

{\mathrm{H}^ + }\;is\;6.0 \times {10^5}\;\mathrm{mol}/\mathrm{L} \times \mathrm{s},

how long will it take for

{\mathrm{H}^ + }

to disappear from the drop

View answer and explanation

Correct option: B

2.0 \times {10^{ - 8}}\;\mathrm{s}

Option B (

2.0 \times {10^{ - 8}}\;\mathrm{s}

) is correct because it is the time calculated by dividing the total moles of H+ by the rate of disappearance after converting the rate to moles/second.

NEET UG CHEMISTRY MCQ: CHEMICAL KINETICS

More Questions from CHEMICAL KINETICS