(B) The Avogadro number is $N_A = 6.023 \times 10^{23}$.Time taken in seconds $= \frac{6.023 \times 10^{23}}{10^{20}} = 6023 \text{ seconds}$.To conve

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(B) The Avogadro number is $N_A = 6.023 \times 10^{23}$.Time taken in seconds $= \frac{6.023 \times 10^{23}}{10^{20}} = 6023 \text{ seconds}$.To convert seconds into hours,divide by $3600$:Time in hours $= \frac{6023}{3600} \approx 1.673 \text{ hours}$.

Class 11 Chemistry Some Basic Concepts of Chemistry The mole concept

Medium

If $10^{20}$ grains are distributed in one second,then the time taken to distribute an Avogadro number of grains is ........ hours.

A
$0.1673$
B
$1.673$
C
$16.73$
D
$167.3$

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Some Basic Concepts of Chemistry

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The mole concept

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$0.1$ mole of compound $'S'$ will weigh $.......... \ g$.
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The mass of $10^{21}$ atoms of $Cu$ is ..... grams.

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If $10^{20}$ grains are distributed in one second,then the time taken to distribute an Avogadro number of grains is ........ hours.

Similar Questions

If $19.7 \ kg$ of gold is confiscated from a thief,how many atoms of gold were confiscated from him? $(Au = 197 \ g/mol)$

How many moles of helium gas occupy $22.4 \ L$ at $0^o C$ and $1 \ atm$ pressure?

$A$ mixture has $18 \ g$ of water and $414 \ g$ of ethanol. The mole fraction of water in the mixture is (assume ideal behavior of the mixture):

$0.1$ mole of compound $'S'$ will weigh $.......... \ g$.(Given molar mass in $g \ mol^{-1} \ C: 12, H: 1, O: 16$)

The mass of $10^{21}$ atoms of $Cu$ is ..... grams.

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$0.1$ mole of compound $'S'$ will weigh $.......... \ g$.
(Given molar mass in $g \ mol^{-1} \ C: 12, H: 1, O: 16$)