Answer:
Sol: Information shadow: In this
problem some obvious facts have to be used, i.e.,
Mass of single electron \[({{m}_{e}})=9.1\times
{{10}^{-28}}g\]
\[=9.1\times
{{10}^{-31}}kg\,\,\,...(i)\]
Problem solving strategy:
In this problem, it is to calculate that how many electrons will together weigh
1 g.
Let x electrons will together
weigh 1 g.
\[\therefore
\,\,\,\,\,\,\,\,\,\,\,\,\,x\times {{m}_{e}}=1g\]
Working it out: From
eqns. (ii) and (i)
\[x\times 9.1\times
{{10}^{-28}}=1\]
\[x=\frac{1}{9.1\times
{{10}^{-28}}}1.098\times {{10}^{27}}\]
(ii) Calculate the mass and charge of
one mole of electrons.
Sol: Information shadow:
In the present problem only the name of
particle is given, i.e., electron.
Mass of an electron \[{{m}_{e}}=9.1\times
{{10}^{-31}}kg\]
Charge of an electron \[{{q}_{e}}=1.6\times
{{10}^{-19}}C\]
Problem solving strategy:
One mole is a collection of \[6.023\times
{{10}^{23}}\]particles
\[\therefore \]Mass of 1 mole
electrons =
\[6.023\times {{10}^{23}}\times
{{m}_{e}}\,\,\,\,\,\,\,\,\,\,\,....(i)\]
Charge of 1 mole electrons =
\[6.023\times {{10}^{23}}\times
{{q}_{e}}\,\,\,\,\,\,\,\,\,\,\,....(ii)\]
Working it out:
From Eqn. (i): Mass of 1 mole
electrons
\[=6.023\times {{10}^{23}}\times
9.1\times {{10}^{-31}}\]
\[=5.48\times {{10}^{-7}}kg\]
From Eqn. (ii): Charge of 1 mole
electrons
\[=6.023\times {{10}^{23}}\times
1.6\times {{10}^{-19}}\]
\[=96368C\]
DO YOU KNOW?
Charge of 1 mole electron = 1
faraday
1 faraday = 96500 C (Approximately)
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