A drop of solution volume -0-55 ml contains 3 - 10-2 mole of H - -If the rate constant of disappearance of H -is 1 - 102 mole per litre per minute- How long in minutes will it take for all the H -ions in a litre of the solution to disappear-

0.05 ml has 3× 10^ 2 mole of H^+ 1000 ml has 3× 10^ 2× 1000/0.05=600 moles of H^+ k=x/t (By the units the reaction is of 0th order) t=600/1× 10^2=6 minutes. ...

A drop of sol...

Question

A drop of solution (volume = 0.55 ml) contains $3 \times 10^{- 2}$ mole of $H^{+}$ . If the rate constant of disappearance of $H^{+}$ is $1 \times 10^{2}$ mole per litre per minute. How long (in minutes) will it take for all the $H^{+}$ ions in a litre of the solution to disappear?

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Solution

0.05 ml has 3 $\times 10^{- 2}$ mole of $H^{+}$

1000 ml has $\frac{3 \times 10^{- 2} \times 1000}{0.05} = 600$ moles of $H^{+}$

$k = \frac{x}{t}$ (By the units the reaction is of 0th order)

$t = \frac{600}{1 \times 10^{2}} = 6$ minutes.

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A drop of solution (volume = 0.55 ml) contains 3×10−2 mole of H+. If the rate constant of disappearance of H+ is 1×102 mole per litre per minute. How long (in minutes) will it take for all the H+ ions in a litre of the solution to disappear?

0.05 ml has 3×10−2 mole of H+ 1000 ml has 3×10−2×10000.05=600 moles of H+ k=xt (By the units the reaction is of 0th order) t=6001×102=6 minutes.

A drop of solution (volume = 0.55 ml) contains $3 \times 10^{- 2}$ mole of $H^{+}$ . If the rate constant of disappearance of $H^{+}$ is $1 \times 10^{2}$ mole per litre per minute. How long (in minutes) will it take for all the $H^{+}$ ions in a litre of the solution to disappear?

0.05 ml has 3 $\times 10^{- 2}$ mole of $H^{+}$

1000 ml has $\frac{3 \times 10^{- 2} \times 1000}{0.05} = 600$ moles of $H^{+}$

$k = \frac{x}{t}$ (By the units the reaction is of 0th order)

$t = \frac{600}{1 \times 10^{2}} = 6$ minutes.