A drop of solution volume - 0-10ml contains 6-10-6 moles of H- if the rate constant of disappearence of H- is 1-107 mole litre-1 sec-1- The time taken for H- in drop to disappear is -

The correct option is D 6×10^ 9sec 6×10^ 6 moles in 0.10 ml corresponds to 6×10^ 6 moles0.101000=6×10^ 2 moles .In one second, 1×10^ ...

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Byju's Answer

Standard VII

Chemistry

Balanced Chemical Equation

A drop of sol...

Question

A drop of solution (volume = 0.10ml) contains $6 \times 10^{- 6}$ moles of $H^{+}$ , if the rate constant of disappearence of $H^{+}$ is $1 \times 10^{7}$ mole $l i t r e^{- 1} s e c^{- 1}$ . The time taken for $H^{+}$ in drop to disappear is :

6 \times 10^{- 9} s e c

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7 \times 10^{- 9} s e c

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12 \times 10^{- 9} s e c

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18 \times 10^{- 9} s e c

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Solution

The correct option is D $6 \times 10^{- 9} s e c$
$6 \times 10^{- 6}$ moles in $0.10$ ml corresponds to $\frac{6 \times 10^{- 6} m o l e s}{\frac{0.10}{1000}} = 6 \times 10^{- 2} m o l e s$ .
In one second, $1 \times 10^{7}$ moles disappear.
Hence, the time required for the disappearance of $6 \times 10^{- 2}$ moles is $\frac{6 \times 10^{- 2}}{1 \times 10^{7}} = 6 \times 10^{- 9} s e c$ .

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A drop of solution (volume = 0.10ml) contains 6×10−6 moles of H+, if the rate constant of disappearence of H+ is 1×107 mole litre−1 sec−1. The time taken for H+ in drop to disappear is :

The correct option is D 6×10−9sec6×10−6 moles in 0.10 ml corresponds to 6×10−6moles0.101000=6×10−2moles.In one second, 1×107 moles disappear.Hence, the time required for the disappearance of 6×10−2 moles is 6×10−21×107=6×10−9sec.

A drop of solution (volume = 0.10ml) contains $6 \times 10^{- 6}$ moles of $H^{+}$ , if the rate constant of disappearence of $H^{+}$ is $1 \times 10^{7}$ mole $l i t r e^{- 1} s e c^{- 1}$ . The time taken for $H^{+}$ in drop to disappear is :